Jay R. Yablon
Nov4-06, 03:32 PM
[Moderator's note: Please try to wrap your lines to 72 characters, as I
have done below. This allows the post to be quoted 4 times with a
standard symbol like "> " without exceeding the 80-character limit,
which is what many displays have. -P.H.]
Dear SPR folks:
In QED, one transforms a Dirac wavefunction psi according to psi -->
psi' = e^ia psi, where a is a local gauge parameter. As a consequence
of d_u (understood to be a partial derivative) operating on psi in the
Dirac equation (gamma^u d_u - m) psi, to maintain gauge invariance we
are forced to absorb d_u a by introducing a vector field A^u
transforming as eA_u --> eA_u' + d_u a, where A_u is the vector
potential (field) for the photon and e is the running electric charge
(such that a = (1/4pi)e^2 ~ 1/137.036 at low probe energy). There are,
if you will, two levels to this gauge transformation: the e^ia which
multiplies the wavefunction psi, and the eA_u --> eA_u' + d_ua which
describes how the photon transforms.
In the *linear approximation* of general relativity, the linear field
equation kappa_0 T^uv = d^a d_a (phi^uv) in said to be invariant under
the gauge transformation h_uv --> h_uv' = h_uv + d_u /\^v + d+v /\^u
where phi^uv = h^uv - (1/2) n^uv h represents the graviton, where h^uv
is the gravitational field tensor, and where n^uv is the Minkowski
metric tensor. This, of course, is analogous to eA_u --> eA_u' + d_ua,
but the gauge parameter /\^v is a vector field rather than the scalar
field a of QED.
My question is this: what is the general relativistic analog, if any, to
the wavefunction psi in QED, and *especially*, to the term e^ia which
multiplies psi in QED? If we were to write down an e^i/\^u and use this
to transform some type of wavefunction PSI (of unknown structure for
now, capitalized to differentiate from the psi of QED), that is, if we
were to write down e^i/\^u PSI, this would automatically add a vector
index to whatever the structure is of PSI. I have played with a variety
of structures for putting the gauge parameter /\^u into a complex
exponential and for the associated PSI, and it is not clear to me
whether the QED psi --> psi' = e^ia psi even has an analogy in general
relativity, even in the linear approximation.
I note also that in non-Abelian gauge theory (QCD for example), we use
psi --> psi' = e^i lambda_j a_j psi where j=1,2,3...8 for QCD is an
internal symmetry index and lambda_j are the QCD structure matrices.
Here, the e^i lambda_j a_j is a 3x3 second rank SU(3) tensor and the psi
accordingly becomes an SU(3) vector (with four-component Dirac structure
implied). Even with this analogy, it is not clear how to structure PSI
or a complex exponential containing /\^u. It seems that e^i/\^u needs
to be made into a second rank (spacetime) tensor and PSI into a
(spacetime) vector to maintain the structure of PSI (which we can do by
an index summation), but that would seems to mess up the transformation
h_uv --> h_uv' = h_uv + d_u /\^v + d+v /\^u.
Can someone please elucidate how this all hangs together, if this is
known?
Thanks.
Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
have done below. This allows the post to be quoted 4 times with a
standard symbol like "> " without exceeding the 80-character limit,
which is what many displays have. -P.H.]
Dear SPR folks:
In QED, one transforms a Dirac wavefunction psi according to psi -->
psi' = e^ia psi, where a is a local gauge parameter. As a consequence
of d_u (understood to be a partial derivative) operating on psi in the
Dirac equation (gamma^u d_u - m) psi, to maintain gauge invariance we
are forced to absorb d_u a by introducing a vector field A^u
transforming as eA_u --> eA_u' + d_u a, where A_u is the vector
potential (field) for the photon and e is the running electric charge
(such that a = (1/4pi)e^2 ~ 1/137.036 at low probe energy). There are,
if you will, two levels to this gauge transformation: the e^ia which
multiplies the wavefunction psi, and the eA_u --> eA_u' + d_ua which
describes how the photon transforms.
In the *linear approximation* of general relativity, the linear field
equation kappa_0 T^uv = d^a d_a (phi^uv) in said to be invariant under
the gauge transformation h_uv --> h_uv' = h_uv + d_u /\^v + d+v /\^u
where phi^uv = h^uv - (1/2) n^uv h represents the graviton, where h^uv
is the gravitational field tensor, and where n^uv is the Minkowski
metric tensor. This, of course, is analogous to eA_u --> eA_u' + d_ua,
but the gauge parameter /\^v is a vector field rather than the scalar
field a of QED.
My question is this: what is the general relativistic analog, if any, to
the wavefunction psi in QED, and *especially*, to the term e^ia which
multiplies psi in QED? If we were to write down an e^i/\^u and use this
to transform some type of wavefunction PSI (of unknown structure for
now, capitalized to differentiate from the psi of QED), that is, if we
were to write down e^i/\^u PSI, this would automatically add a vector
index to whatever the structure is of PSI. I have played with a variety
of structures for putting the gauge parameter /\^u into a complex
exponential and for the associated PSI, and it is not clear to me
whether the QED psi --> psi' = e^ia psi even has an analogy in general
relativity, even in the linear approximation.
I note also that in non-Abelian gauge theory (QCD for example), we use
psi --> psi' = e^i lambda_j a_j psi where j=1,2,3...8 for QCD is an
internal symmetry index and lambda_j are the QCD structure matrices.
Here, the e^i lambda_j a_j is a 3x3 second rank SU(3) tensor and the psi
accordingly becomes an SU(3) vector (with four-component Dirac structure
implied). Even with this analogy, it is not clear how to structure PSI
or a complex exponential containing /\^u. It seems that e^i/\^u needs
to be made into a second rank (spacetime) tensor and PSI into a
(spacetime) vector to maintain the structure of PSI (which we can do by
an index summation), but that would seems to mess up the transformation
h_uv --> h_uv' = h_uv + d_u /\^v + d+v /\^u.
Can someone please elucidate how this all hangs together, if this is
known?
Thanks.
Jay.
_____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com