Re: Relation of g^uv = (1/2) {gamma^u,gamma^v} to gravitational fields

"Igor Khavkine" <igor.kh@xxxxxxxxx> wrote in message
news:1158091928.514623.68050@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

But why do you think that the anticommutators of the corrected
vertex factors Gamma^u will have anticommutators proportional to the
Clifford identity? That's a necessary condition for them to generate a
Clifford algebra.

Is there any reason why they will *not* have anticommutators
proportional to the Clifford identity? Or, are you just saying that
there is no particular reason why they *should*, nor any reason why they
should not? I will ask this same question more precisely, below, in
response to the Lagrangian density you have laid out.

Also, gravity is more than just a metric tensor. It is the metric
tensor coupled to matter in a specific way. For instance, in a
Lagrangian density with some scalar and and spinor fields, it enters
as
follows (G - metric, @ - partial derivative, g - gamma matrix, A -
vector potential):

magnetic moment correction --+
v
L = G @phi @phi + psi g @psi + psi (g + lambda) A psi + ...
^ ^ ^
|__ gravity |__ gravity |__ gravity

When writing down the effective action (which is the only place where
the corrected vertex factor you refer to actually makes an
appearance),
each coefficient shown above changes in a particular way compared to
its bare value. The G, g (in the psi-psi term), and g (in the
psi-A-psi) term will all change according to different rules. There is
no reason to expect them to conspire to change such that G stays equal
to the anticommutators of the g+lambda components.

Igor

OK, your Langrangian density helps to frame the question:

It drives me crazy to try to read equations with ^ _ @ -- etc. So, at
the link below is what I take to be your Lagrangian density written as
an easy-to-read equation:

http://home.nycap.rr.com/jry/Papers/Equation%201.jpg (1)

Now, my question is this: Start in flat spacetime with a Lagrangian
given by:

http://home.nycap.rr.com/jry/Papers/Equation%202.jpg (2)

Then introduce a perturbatively-corrected vertex factor Lambda^u so that
gamma^u --> gamma^u + Lambda^u. Is there any reason why the corrected
Lagrangian ought *not* look like the following?

http://home.nycap.rr.com/jry/Papers/Equation%203.jpg (3)

Or, are you just saying there is no reason why it should look like this
any more than it should not look like this, AND that the question of
whether the anticommutators of the corrected vertex factors Gamma^u will
have anticommutators proportional to the Clifford identity is what will
determine whether or not one ends up with a Lagrangian that looks like
(3)?

Thank you.

Jay.




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