What is the Gravitationally-Covariant derivative of the coordinates x^u?
Jay R. Yablon
gravitationally-covariant derivative of the coordinates x^u. The
context for the question I am raising here, arises from a comment made
by Mark Hopkins in a separate thread on sci.physics.research
(Integration by Parts of the Maxwell QED Action, Tuesday, October 06,
2009 6:07 PM) that "There is no known formulation of path integrals to
curved spacetimes." I have since been thinking quite a lot about this
comment, and asking myself whether there might be a way to formulate
path integrals in curved space time and, if so, how one might do so.
Part of these considerations, leads me to the question posed in the
title of this thread.
To provide the context in which this question arises, consider the
path integration of the free Maxwell action given in flat spacetime by
(& == ordinary partial derivative, $ == integral):
S = boundary term +
$ d^4x [.5 A_u [g^uv (&_s &^s + m^2) - &^v &^u] A_v + J^v A_v] (1)
To use this in a path integral, one crucial step is to define the
spacetime propagator D_av(x-y) as the inverse of the operator in the
middle of (1), according to:
D_av(x-y) [g^uv (&_s &^s + m^2) - &^v &^u] == delta^u_a delta^(4)(x-y)
(2)
where delta^u_a is the Kronecker unit matrix and delta^(4)(x-y) is the
Dirac delta. The Dirac delta may be specified as the integral over
momentum space of the Fourier kernel exp[i k_s (x-y)^s]. When we also
Fourier transform D_av(x-y) into momentum space, we find that:
(2pi)^4 D_av(x-y) [g^uv (&_s &^s + m^2) - &^v &^u]
= $d^4k [g^uv (&_s &^s + m^2) - &^v &^u] exp[i k_t x^t] (3)
= $d^4k [g^uv (- k_s k^s + m^2) + k^v k^u] exp[i k_t x^t]
In the final line of the above, we have implicitly performed the
calculation:
&_s exp[i k_t x^t] = i &_s [k_t x^t] exp[i k_t x^t]
= i &_s [k_t x^t] exp[i k_t x^t] = i delta^t_s k_t exp[i k_t x^t]
(4)
= i k_s exp[i k_t x^t]
which includes:
&_s x^t = delta^t_s (5)
Equation (4) shows the origin (or an least *an* origin) of the fabled
quantum mechanical substitution:
&_s --> i k_s (6)
Equation (5) says that "the ordinary partial derivative &_s of the
coordinates x^t, is the Kronecker delta^t_s." If x^t=(t,x,y,z) are the
coordinates, each regarded as independent, then, for example, &t/&t = 1
and &x/&x = 1 say that the time coordinate varies identically with
itself, and the space coordinates vary identically with themselves.
And, for example, &x/&y = 0 says that, *a priori*, x and y are
independent coordinates, i.e., that a change in x tells us precisely
nothing about how y changes.
Now, in curved spacetime, ordinary derivatives &_u become
gravitationally-covariant derivatives which I will denote by &;_u. For
example, for a lower-indexed four-vector A_v, the covariant derivative
is:
&;_u A_v = &_u A_v - Gamma ^b_uv A_b (7)
where Gamma ^b_uv are the Christoffel connections. And, of course,
for the metric tensor, we have the special metricity relationship:
&;_s g_uv = 0 (8)
But, what happens when we take the *covariant derivative* of the
coordinates x^u? Specifically, this question arises because, if we wish
to consider (1) above in curved spacetime, then at the very least, we
must promote all partial derivatives to *covariant* derivatives and thus
rewrite (1) as:
S = boundary term +
$ d^4x [.5 A_u [g^uv (&;_s &;^s + m^2) - &;^v &;^u] A_v + J^v A_v] (9)
We also note that the Riemann curvature tensor may be *defined* as a
measure of the non-commutativity [A,B] /= 0 of the covariant derivative,
i.e., that:
R^a_buv A_a == [&;_u &;_v] A_b (10)
and so may be easily be introduced directly into (9) by commuting the
indexes in the term &;^v &;^u A_v. In fact, I often find it helpful to
think of curved spacetime as simply spacetime in which partial
derivatives are non-commuting according to the relationship (10), and so
have been approaching Mark's statement that "There is no known
formulation of path integrals to curved spacetimes" by trying to
calculate the path integral based on (1), but treating the partial
derivatives as *non-commuting* as in (9) and (10).
Now, in curved spacetime, it seems that (2) above will be promoted
to include non-commuting covariant derivatives, thus:
D_av(x-y) [g^uv (&;_s &;^s + m^2) - &;^v &;^u] == delta^u_a
delta^(4)(x-y) (10)
and, that the first two lines of (3) above will become:
(2pi)^4 D_av(x-y) [g^uv (&;_s &;^s + m^2) - &;^v &;^u]
= $d^4k [g^uv (&;_s &;^s + m^2) - &;^v ;&^u] exp[i k_t x^t]
(11)
And, that, brings us to where my question in the thread title arises.
To take the next step with (11), we must know, "What is the
Gravitationally-Covariant derivative of the coordinates x^u?"
Specifically, working from (5) above, (11) above requires us to
calculate:
&;_s exp[i k_t x^t] = i &;_s [k_t x^t] exp[i k_t x^t]
= i &;_s [k_t x^t] exp[i k_t x^t] (12)
which now includes the need to calculate:
&;_u x^v = ??? (13)
I have thought about this in several ways, but am not sure which
one, if any, is correct. First, x^v are the coordinates, and do *not*
form a vector under general coordinate transformations. Thus, it seems
to be misguided to simply use the usual expression for the covariant
derivative of a vector (7) and write:
&;_u x^v = &_u x^v + Gamma ^v_us x^s = delta^v_u + Gamma ^v_us x^s
(14)
I was first inclined to think that since coordinates "vary with
themselves," the correct answer is that:
&;_u x^v = &_u x^v = delta^v_u , (15)
period! In other words, my first thought was that that in curved
spacetime, the covariant derivative of the coordinates should remain
equal to the unit matrix, come hell or high water. But then, I got to
thinking about the vierbein formalism in which the vierbein:
V^a_u = (&y^a (_x) /&x^u) at X = x (16)
That is, we have orthonormal coordinates y^a (_x) (and Lorentz indexes
a) at each spacetime event X = x, as well as world coordinates x^u (and
world indexes u). In the transition from flat to curved spacetime, we
are also moving over to a domain in which we may specify local
coordinates which are tangent to the world coordinates at each event in
the coordinate system. Then, y^a represents the "flat" spacetime
coordinates and x^u represents the world coordinates, and, perhaps, in
(13) and (15), we should be using both types of coordinate and so
rewrite (12) with x-->y in the Fourier kernel as:
&;_s exp[i k_t y^t] = i &;_s [k_t y^t] exp[i k_t y^t]
= i &;_s [k_t y^t] exp[i k_t y^t] (17)
wherein the answer to the thread question (13), cast in the form of (15)
with the world index --> local index transition v-->a, is
*** &;_u y^a = V^a_u . (18) *** ???
That is, might it be that *the vierbein specifies the covariant
derivative of the coordinates x^u* (which we should really think of as
y^u in the sense of vierbein theory)?
To examine the flat spacetime limit, keep in mind that in general,
the metric tensor:
g_uv = V^a_u V^b_v n^ab (19)
where n^ab is the Minkowski metric tensor. Therefore, in the flat
spacetime limit,
V^a_u --> delta^a_u (20)
and so, via (18) and (20):
&;_u y^a --> delta ^a_u . (21)
which is the usual flat spacetime relationship.
Is this correct? If not, how might I be thinking about all of this?
Thanks,
Jay
____________________________
Jay R. Yablon
Email: jya...@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm
Jay R. Yablon
I say "To provide the context in which this question arises, consider
the path integration of the *free* Maxwell action. . ." The word free
should have been omitted. After I first wrote this, I added the source
term J^v A_v to (1), but forgot to take out "free." Jay.
"Jay R. Yablon" <jya...@nycap.rr.com> wrote in message
news:7k15jeF...@mid.individual.net...
> I would like to discuss how one establishes the
> gravitationally-covariant derivative of the coordinates x^u. The
> context for the question I am raising here, arises from a comment made
> by Mark Hopkins in a separate thread on sci.physics.research
> (Integration by Parts of the Maxwell QED Action, Tuesday, October 06,
> 2009 6:07 PM) that "There is no known formulation of path integrals to
> curved spacetimes." I have since been thinking quite a lot about this
> comment, and asking myself whether there might be a way to formulate
> path integrals in curved space time and, if so, how one might do so.
> Part of these considerations, leads me to the question posed in the
> title of this thread.
>
> To provide the context in which this question arises, consider the
> path integration of the free Maxwell action given in flat spacetime by
> (& == ordinary partial derivative, $ == integral):
>
> S = boundary term +
> $ d^4x [.5 A_u [g^uv (&_s &^s + m^2) - &^v &^u] A_v + J^v A_v] (1)
. . .
Tom Roberts
> I would like to discuss how one establishes the
> gravitationally-covariant derivative of the coordinates x^u.
I'm not sure what "gravitationally-covariant derivative" means. But I do
know what the covariant derivative is in differential geometry, and I'll
discuss that.
Geometrically, in an N-dimensional manifold the coordinates {x^u} are a
set of N scalar fields on the manifold, valid in the region of validity
for the coordinate system. They are required to be assigned such that
each field is continuous, and any specific set of N values {x^u} is
unique within the region, specifying a single point of the manifold.
Equivalently, the coordinate system must define a continuous 1-to-1 map
from the region of the manifold to a region of R^N.
So it's clear what the covariant derivative of any one of the {x^u} is:
it's just the covariant derivative of a scalar field. For this to have
meaning, the coordinate fields must of course be at least C^1.
Remarkably, if one differentiates by the {x^u} themselves, one gets a
tensor (this is not obvious):
Dx^u/dx^v = dx^u/dx^v = d^u_v (the usual Kronecker delta)
[The first 3 "d"-s are partials; the fourth "d" is often
written as a lowercase Greek delta.]
NOTE: That first expression LOOKS like the covariant
derivative of a vector x, but it is NOT! -- that's why
obtaining a tensor is not obvious. This is one usage
of component notation that can easily confuse....
In the first = I took advantage of the fact that the covariant
derivative of a scalar field wrt a coordinate is just the ordinary
(partial) derivative wrt the coordinate.
Tom Roberts
Jay R. Yablon
news:G_SdnQeCmP0...@giganews.com...
Thanks Tom,
The result
Dx^u/dx^v = dx^u/dx^v = d^u_v (the usual Kronecker delta)
is one of the possible answers I had considered, and is the simplest
one.
I was wondering if someone might give a second opinion, to confirm that
Tom is correct, since I will be embedding this result in a larger proof
and want to make absolutely certain that this is correct before I
proceed.
Thanks,
Jay