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

Jay R. Yablon wrote:
"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?

Is there any reason to believe that there *aren't* any positive
integers a,b,c that satisfy a^n + b^n = c^n, for n>2? If you are not
familiar with Fermat's Last Theorem and its history, you may think:
"Hey, this is a simple equation, there's got to be some solutions!"
But, as the celebrated proof of Fermat's theorem has shown, there exist
no such a,b,c. The fact is that the above equation is a restrictive
condition on triples of positive integers. Unless you have reason to
believe otherwise, you should never simply assume that any restrictive
condition always has solutions. So, instead of looking for reasons why
such a condition *shouldn't* hold, you should look for reasons why it
*should*.

Getting back on topic, if A and B are generic elements of a Clifford
algebra then their anticommutator {A,B} will be another generic element
of the algebra. And since the subspace of elements proportional to the
identity is only one-dimensional in a 2^n dimensional algebra (n -
space-time dimension), it is very unlikely that {A,B} will be contained
in it.

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.

That's unfortunate, because there is at least one benefit of the ASCII
only nature of Usenet. In case of mathematical notation, it forces you
to distill equations and formulas to their absolute essence to make
them readable. The latter should be the goal of any notation.

[...]

The answer to the your new question is already present in my previous
reply. To help you decode it, I will place emphasis on the key words
that may have obscured its meaning.

... [different terms of] the effective action ... will all change
according to different rules ... compared to [their] bare values.

So, what is an "effective action"? And how do the terms in the
effective action change compared to their bare value?

Finally, the actual answer to your question is negative. Again,
*generically* there is no reason for any of the coefficients of the
effective action to change in unison with the coefficient of
fermion-photon coupling. In fact, the same kind of coefficient will not
change in the same way between different fermions, as evidenced by
say different magnetic moments of different elementary particles.

Igor

.

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