Compilation of various pieces of original remarks and answers
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REMARK:
... the cosmological constant with its usual Planck scale value is present in
Eq. 11 of the manuscript. Then, in going on to Eq. 24, a subtraction is
introduced (the square root of the metric determinant is replaced by the
square root minus one). This constitutes the usual ad hoc subtraction of the
bare cosmological constant.
ANSWER:
... at first sight, the subtraction could seem ad hoc, but it is not quite so.
(i) Standard argument: it follows from generally accepted rules of quantum
field theory, e.g. in the path-integral formalism it corresponds to a
normalization of measure;
(ii) Our argument: the "one" (constant) corresponds to a contribution
without any external lines, i.e. it does not influence external fields by
construction, and therefore it should be excluded.
REMARK:
Following Eq. 24, the expression is expanded about the present time, and
a residual is obtained which vanishes today but which has a similar
time-dependence as that of dark energy. Hence, a connection with dark energy
is postulated. However, the expansion about a fixed time does not make sense.
There is nothing special about the present time which allows a subtraction to
be linked with the present time.
ANSWER:
In principle, we could expand about any fixed time. It is only a technical assumption.
But to obtain present experimental values (e.g. density) we should use present
experimental data (e.g. the present Hubble constant).
REMARK:
The purpose of this paper is to derive the contribution to the vacuum
energy of the universe from the single loop of a free quantum matter field, in
order to achieve better understanding of the cosmological constant problems.
Needless to say that this calculation has been approached previously in many
different ways. Authors adopt a very simple approach and postulate that
the mentioned quantum loop should be cut-off at the Planck scale. Then,
using the Schwinger-DeWitt expansion and taking only the massless scalar
field case they notice that only the a0 coefficient gives contribution to the
proper cosmological constant. Furthermore, they notice that making the
expansion of the metric around the flat background, the first order term
can be eliminated by the coordinate transformation and therefore end up
with the second order term which is a product of the squares of the Hubble
parameter H and the cut-off parameter, that is the Planck mass. As a result
they arrive at the induced cosmological constant which has a ”correct” order
of magnitude.
ANSWER:
Ideologically, our work consists of two parts. The first part presents the
very idea, whereas the second part presents a realization of that idea. We do
not insist that the realization proposed is the best and final one.
REMARK:
Despite the above scheme looks appealing, it has obvious weak points.
For instance, if authors incorporate the field with the mass M, the next
Schwinger-DeWitt term a_1 will produce contribution which will be a product
of the squares of the mass M and the Planck mass. Indeed, even for
neutrino this contribution will be 60 orders of magnitude greater than the
”correct” value. The origin of this occurences is that the scheme suggested
by the authors is not correct from the quantum field theory viewpoint. In
fact, after the effective action is calculated, one can not perform
the expansion around the flat background, because the cosmological constant term
should be covariant, exactly as other terms in the effective action of gravity.
The explanation of the coordinate dependence of the linear in H term is
that the effective action of gravity is covariant and therefore can not include
terms which are odd in metric derivatives. The discussion of this point has
been given recently in gr-qc/0801.0216, where one can also find many other
references on the subject.
ANSWER:
A massive field with the mass M poses a problem. We can propose the
following two independent solutions of that problem: a “theoretical” and a
“technical” one.
The “theoretical” solution would assume, rather a widely accepted idea, that
the mass of a particle is not a fundamental entity but rather a derived one.
In principle, all masses in the standard model are generated via the Higgs
mechanism. According to this point of view we could legitimately assume
that all masses should be neglected at this stage.
The “technical” solution instead would order us to treat the M^2 part
of the a_1 coefficient analogously to the a_0 coefficient. In fact, functional
dependence on metric field, and only metric dependence matters, in the both
terms is identical. The only difference resides in coefficients and
consists in different powers of M and s (UV cutoff), i.e. we have
M^0 s^-2 and M^2 s^-1, respectively. Therefore, actually the contribution
of M^2 part of a_1 would be even less than that of a_0.
REMARK:
... the cosmological constant problem is much deeper than the conflict between
the calculation based on the Planck cut-off and the observable value.
At the first place this problem involves the fine-tuning between the induced
and vacuum components of the cosmological constant, as it is explained,
e.g., in [2].
ANSWER:
Firstly, the huge value of the vacuum energy density is a very annoying
feature in itself, and, it seems, it is desirable to explain it somehow,
independently of the further issue of the cosmological constant.
Secondly, there is an old and a bit controversial idea of induced gravity.
Intriguingly, it appears that it also yields a correct order of magnitude of the
coefficient in front of the Hilbert–Einstein gravitational action. But one of
the arguments against that idea says that the huge value of the cosmological
constant invalidates the very idea, at least in its Sakharov’s version (demand
one-loop dominance in Visser’s classification [7]), i.e. when there are only
induced terms, without bare ones. Now, the tamed cosmological term and
the Hilbert–Einstein gravitational action with the purely induced coefficient
(only) of a correct order of magnitude, both would fit the Sakharov’s version
of induced gravity idea consistently.
