Re: Why

From: Ken S. Tucker (dynamics_at_vianet.on.ca)
Date: 02/20/05 

Date: 20 Feb 2005 04:07:57 -0800

David McAnally wrote:
> Non Ame <noname@nospam.net> writes:
>
> >"Ken S. Tucker" <dynamics@vianet.on.ca> writes:
>
> <snip>
>
> >The Hamiltonian for a single free charged body in an electromagnetic
field
> >is:
>
> > H = (p - q A)^2/(2m) + q phi
>
> >in the non-relativistic case,
>
> > H = sqrt[(p - q A)^2 + m^2] + q phi
>
> >in the relativistic case, and
>
> > H = alpha.(p - q A) + m beta + q phi
>
> >in the case of the Dirac electron.
>
> >Here, A is the vector potential, phi is the scalar potential, q is
the
> >charge, p is the canonical momentum, m is the mass, and alpha and
beta
> >are 4x4 matrices such that
>
> > alpha_i alpha_j + alpha_j alpha_i = 2 delta_{ij}, i, j = 1, 2, 3,
>
> > alpha_i beta + beta alpha_i = 0, i = 1, 2, 3,
>
> > beta^2 = 1.
>
> Note that in the case of the Dirac electron,
>
> (H - q \phi)^2 - (p - q A)^2 = m^2 - \hbar q \Sigma.B,
>
> where
>
> \Sigma_1 = - i \alpha_2 \alpha_3 = i \alpha_3 \alpha_2,
>
> \Sigma_2 = - i \alpha_3 \alpha_1 = i \alpha_1 \alpha_3,
>
> \Sigma_3 = - i \alpha_1 \alpha_2 = i \alpha_2 \alpha_1.
>
> Note that (H - q \phi)^2 - (p - q A)^2 has a dependence on the
magnetic
> field in the case of the Dirac electron.
>
> The Lagrangian is
>
> L = (1/2) m v^2 + q A.v - q \phi
>
> in the non-relativistic case, and
>
> L = - m (1 - v^2)^(1/2) + q A.v - q \phi
>
> in the relativistic case.
>
> Define the one-form
>
> \omega = A.dx - \phi dt = A_x dx + A_y dy + A_z dz - \phi dt,
>
> then \omega is the one-form corresponding to the electromagnetic
> potential.
>
> In the non-relativistic case, Hamilton's Principle of Least Action
states
> that the path taken by the charge from (t_i,x_i,y_i,z_i) to
> (t_f,x_f,y_f,z_f) is the path, P, which minimizes the action,
>
> S = (1/2) m \int_P v^2 dt + q \int_P \omega,
>
> over all paths from (t_i,x_i,y_i,z_i) to (t_f,x_f,y_f,z_f) (or more
> generally, the path is one at which S takes a stationary value).
>
> In the relativistic case, Hamilton's Principle of Least Action states
that
> the path taken by the charge from (t_i,x_i,y_i,z_i) to
(t_f,x_f,y_f,z_f)
> is the path, P, which maximizes
>
> S = m s_P - q \int_P \omega
>
> over all paths from (t_i,x_i,y_i,z_i) to (t_f,x_f,y_f,z_f) (or more
> generally, the path is one at which S takes a stationary value).
Here,
> s_P is the proper time along P, and S is the negative of the action.
>
> This latter statement (the consequence of Hamilton's Principle of
Least
> Action in the relativistic case) has an obvious generalization to
General
> Relativity. Subject to topological considerations, the
electromagnetic
> potential is realizable in General Relativity as a one-form and so
the
> path, P, taken by a charge from one event E_i to a later event E_f is
the
> path which maximizes
>
> S = m s_P - q \int_P \omega
>
> over all paths from E_i to E_f, where s_P is the proper time along P.
>
> In the non-relativistic case, the difference between S_P and S_Q for
paths
> P and Q is
>
> (m/2) (\int_P v^2 dt - \int_Q v^2 dt) + q (\int_P \omega - \int_Q
\omega).
>
> Since paths P and Q have the same endpoints, then P - Q is a closed
curve,
> and so, by Stokes' Theorem, the second term is equal to q \int_M F,
where
> M is a surface bounded by P - Q, and F = d\omega, so that
>
> F = E_x dx dt + E_y dy dt + E_z dz dt + B_x dy dz + B_y dz dx + B_z
dx dy
>
> (recall that dx dt = - dt dx, etc, as these are two-forms). Note
also
> that dF = 0 (a necessary consequence of F = d\omega) is just
Maxwell's
> homogeneous equations. It follows that the difference between S_P
and
> S_Q for paths P and Q is
>
> (m/2) (\int_P v^2 dt - \int_Q v^2 dt) + q \int_M F,
>
> and so the path taken by the charge depends only on the
electromagnetic
> field along the path, and on no other information about the
> electromagnetic potential.
>
> In the case of Special Relativity (and also of General Relativity),
the
> difference between S_P and S_Q for paths P and Q is
>
> m (s_P - s_Q) - q (\int_P \omega - \int_Q \omega).
>
> Since paths P and Q have the same endpoints, then P - Q is a closed
curve,
> and so, by Stokes' Theorem (subject to topological considerations in
the
> case of General Relativity), the second term is equal to q \int_M F,
where
> M is a surface bounded by P - Q, and F = d\omega. It follows that
the
> difference between S_P and S_Q for paths P and Q is
>
> m (s_P - s_Q) - q \int_M F,
>
> and so the path taken by the charge depends only on the
electromagnetic
> field along the path, and on no other information about the
> electromagnetic potential. This is also true in the case of General
> Relativity, independently of topological considerations, since P and
Q can
> be taken to be neighbouring curves, and there is guaranteed to be a
> surface bounded by P - Q.
>
> The fact that \int_M F is independent of the choice of surface M with

> boundary P - Q is evident from Stokes' Theorem. Alternatively, let M
and
> N be surfaces with boundary P - Q, then M - N is a closed surface,
and
> therefore (subject to topological consiferations in the case of
General
> Relativity) M - N is the boundary of a three-dimensional manifold V,
and
> so, by Stokes' Theorem,
>
> \int_M F - \int_N F = \int_V dF = 0,
>
> since dF = 0, and so \int_M F = \int_N F.
>
> For a gauge transformation, A' = A + grad \chi and \phi' = \phi -
d\chi/dt
> for some function \chi(t,x,y,z). It follows that
>
> \omega' = \omega + d\chi,
>
> and d\chi is an exact one-form.
>
> In the non-relativistic case,
>
> S'_P - S_P = q [\chi(t_f,x_f,y_f,z_f) - \chi(t_i,x_i,y_i,z_i)],
>
> so that the actions along all paths are changed the same amount (i.e.
the
> amount of change is the same for all paths, since the change is
dependent
> only on the endpoints). Since the path that minimizes S also
minimizes
> S', then the optimal path is independent of the choice of gauge, as
was to
> be expected for the non-relativistic case, since the path depends
only on
> the electromagnetic field.
>
> Similar considerations hold in the relativistic case, and in the case
of
> General Relativity.
>
> In the context in which the electromagnetic potential, \omega, is a
> connection and the electromagnetic field. F, is the curvature,
Maxwell's
> homogeoneus equation, dF = 0, is just the Bianchi indentity.
>
> <snip>
>
> >>I think real good, GR's G_uv=T_uv is damn
>
> Does this perhaps explain the reason why Ken S. Tucker responded to
my
> observation that
>
> :::The reality is that the time dilation within the context of
General
> :::Relativity is related to the gravitational potential within the
context
> :::of Newtonian mechanics. The relation between the time dilation in
the
> :::context of General Relativity and the gravitational potential
within
> :::the context of Newtonian mechanics should be established fairly
early
> :::on in the derivation of G = 8 pi T in any book on General
Relativity.
>
> by describing it as BS. Ken, while it is true that you can rewrite
the
> equation G_{uv} = k T_{uv} as G_{uv} = T_{uv} by setting k equal to
1,
> that does not eliminate the problem. It only transfers the problem
to a
> question of conversion factor. We know that one second is equal to
> 299 792 458 metres (upon setting c = 1), for example. Your setting k

> equal to one gives us a very specific conversion factor between
kilograms
> and metres, and specifies a certain force as being equal to unity
> (i.e. it gives a specific numerical value for 1 Newton).
>
> Once you have set k equal to 1, how do you know the value of Newton's

> gravitational constant, G? In the Newtonian approximation, your
> specification of the constant k requires that the gravitational force

> between two masses be attractive with a strength of m M/(8 \pi r^2),
> where m and M are the masses, and r is the distance between them, and
the
> gravitational potential energy is U = - m M/(8 \pi r). Similarly,
under
> your convention, Ken, the Schwarzchild metric is
>
> ds^2 = (1 - M/(4 \pi r)) dt^2 - (1 - M/(4 \pi r))^(-1) dr^2 - r^2
d\theta^2
>
> - r^2 sin^2(\theta) d\phi^2.
>
> We know this because comparison with the Newtonian approximation
yields
> that the numerical value of G is 1/(8 \pi).
>
> As I commented, we also require the Newtonian approximation to
determine
> the conversion factor between kilograms and metres (and to calculate
the
> mass of the Sun and the mass of the Earth in kilograms, as opposed to
in
> metres), and also a numerical value of one Newton, under your
convention,
> with the result that
>
> one metre is approximately 5.36*10^25 kilograms,
>
> the numerical value of one Newton is approximately
2.075*10^(-43).
>
> Cavendish's experiment, which was set up to determine Newton's
> gravitational constant, G, loses none of its value as a result of
your
> convention, Ken, since your convention merely changes it into a
> measurement of the conversion factor between mass and length (and
also
> a measurement of a numerical value for the unit of force).
>
> David
>
> -----

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