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| Jay R. Yablon... |
 Posted: Sun Oct 18, 2009 10:11 am |
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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 (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 . (1 *** ???
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 (1 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: jyablon at (no spam) nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.roadrunner.com/~jry/FermionMass.htm |
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| Jay R. Yablon... |
| Posted: Sun Oct 18, 2009 10:57 pm |
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There was one minor error in my original post. In the second paragraph,
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" <jyablon at (no spam) nycap.rr.com> wrote in message
news:7k15ctF3815t7U1 at (no spam) mid.individual.net...
[quote]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 (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)
.. . .[/quote] |
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| Tom Roberts... |
| Posted: Tue Oct 20, 2009 5:45 am |
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Jay R. Yablon wrote:
[quote]I would like to discuss how one establishes the
gravitationally-covariant derivative of the coordinates x^u.
[/quote]
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 |
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| Jay R. Yablon... |
| Posted: Tue Oct 20, 2009 12:26 pm |
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"Tom Roberts" <tjroberts137 at (no spam) sbcglobal.net> wrote in message
news:G_SdnQSCmP2tUUDX4p2dnAA at (no spam) giganews.com...
[quote]Jay R. Yablon wrote:
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
Thanks Tom,[/quote]
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 |
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| ... |
| Posted: Fri Oct 23, 2009 2:24 am |
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Jay R. Yablon <jyablon at (no spam) nycap.rr.com> wrote:
[...]
[quote]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
[/quote]
This is correct. You have to remember, though, that as Tom
emphasized the object x^u is *not* a vector, but rather a
collection of four scalars. This means, for example, that if V_u
is a cotangent vector, the "contraction" X^uV_u doesn't mean
anything -- it's a particular linear combination of the components
of a vector in a particular basis, and will be something essentially
unrelated in a different basis.
Steve Carlip |
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| Jay R. Yablon... |
| Posted: Mon Oct 26, 2009 12:32 pm |
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<carlip-nospam at (no spam) physics.ucdavis.edu> wrote in message
news:hbqf6j$14s$1 at (no spam) skeeter.ucdavis.edu...
[quote]Jay R. Yablon <jyablon at (no spam) nycap.rr.com> wrote:
[...]
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
This is correct. You have to remember, though, that as Tom
emphasized the object x^u is *not* a vector, but rather a
collection of four scalars. This means, for example, that if V_u
is a cotangent vector, the "contraction" X^uV_u doesn't mean
anything -- it's a particular linear combination of the components
of a vector in a particular basis, and will be something essentially
unrelated in a different basis.
Steve Carlip
I understand, and agree completely. Which now raises another question:[/quote]
Let us in fact do what you mention, and in particular, form the term:
exp -i[p_u x^u] (1)
where p_u is a cotangent (energy-momentum) vector. This is of course is
the Fourier kernel and is used in specifying, for example, fermion
wavefunctions psi=exp [p_u x^u]u(p), and, of course, any time we do a
Fourier transform.
If we then subject both p_u and x^u to a general coordinate
transformation, the p_u --> p'_u will of course follow the
transformation law for a covariant (lower-indexed) four vector, but the
x^u will transform instead as:
x^u --> x'^u = x^u - /\^u (x^u) (2)
where /\^u (x^u) is a quadruplet parameter characterizing the
transformation. Therefore, we will have:
p_u x^u --> p_u x^u - alpha(p_u , x^u) (3)
and
exp -i[p_u x^u] --> exp i[alpha(p_u , x^u)] exp -i[p_u x^u] (4)
where alpha(p_u , x^u) is a singlet parameter which is an arbitrary,
local function of p_u and x^u -- arbitrary in the same way and to the
same degree that the choice of coordinates is arbitrary. Explicitly
(d=partial derivative):
alpha(p_u , x^u) = p_u /\^u - p_u d'_s /\^u (x^s -/\^s) (5)
But, if you look closely at (4), you see that although this emerges from
a general coordinate transformation, this resembles a *gauge*
transformation by exp [i alpha].
So, back to your statement "the 'contraction' X^uV_u doesn't mean
anything -- it's a particular linear combination of the components of a
vector in a particular basis, and will be something essentially
unrelated in a different basis." What you are saying, specifically
takes the mathematical form of (4) and (5).
I am wondering (and was studying earlier this year) whether, in the
context of gauge theory, we can essentially "gauge out" the
"meaningless" aspect of this contraction using the gauge symmetry of a
gauge theory? And, whether this is indicative of a deeper connection
between general coordinate invariance and gauge symmetry whereby the two
are really one and gauge symmetry is in some sense a direct consequence
of coordinate symmetry? The main difference -- and it may be a
distinction without a difference -- is that in gauge theory, we can
select the gauge / phase parameter alpha on a completely arbitrary local
basis, whereas here, alpha is completely arbitrary as well, but its
arbitrariness is precisely constrained by the particular arbitrary local
coordinate transformation that we choose to make.
Thanks,
Jay |
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