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Covariant Derivative of a 1-Form
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We worked out what the covariant derivative of a vector must be
here. Now we're going to try
to see, visually, how the covariant derivative of a 1-form must relate
to the coordinate derivative of the 1-form.
The Covariant Derivative of a 1-Form
Again, we want to find the difference between the coordinate
(directional) derivative of a 1-form in a particular coordinate system,
and the
coordinate derivative, in the same direction, evaluated in flat space.
Imagine a 1-form field, ω(P), with
zero derivative at each
point
in flat space. In other words, the 1-form ω is
"really"
the same at every point on the manifold, even though it may appear to
vary from place to place in some particular coordinate system,
S.
We want to find the
directional derivative evaluated using the
S's basis, which
we'll call
B. The
difference we're looking for, which is the difference between the
covariant derivative and the coordinate derivative, will be the value
of that coordinate derivative, negated.
We'll
start by looking at the derivative in the direction of basis vector
eβ.
The α component of the 1-form is the value of the 1-form applied to
basis vector
eα.
The α component of the derivative is, therefore, the derivative of the
value of the
1-form applied to the basis vector
eα, taken along
the
vector
eβ. So we will look at how ω
ξeαξ
changes as we move in the β direction.
Figure 1:
In Figure 1, basis vector
eα is shown before and
after we move one unit in the direction of
eβ.
Basis vector
eα has stretched
and
rotated in the α and γ directions. Studying the diagram should
make clear what the coordinate derivative must be; the following text
is a description in words and equations of what's going on.
The blue lines are the 1-form (see
1-forms
for notes on their representation). The number of blue lines
crossed by any vector is the value of the 1-form applied to that vector.
Since the vector shown is the α basis vector, the number of blue lines
it crosses is the α component of the 1-form.
The
difference in the number of lines crossed by the red versus
gray basis vectors is the change in the α component of the
1-form. The number of blue lines crossed by the thin gray line
labeled "Net Change in
eα" is the change in the α
component of ω.
The green lines show the change in the component value broken down,
itself, into components. The basis vector actually stretched in
the α
direction, and then grew in the γ direction. The total change in
its
value is clearly the sum of those two changes (as we can see from the
picture).
That's it -- the rest is just details! But they're details
we do
need
to fill in, so let us soldier on.
In English, to reiterate, the diagram shows us what the value of the
coordinate derivative with respect to x
β must be,
when the basis vectors are changing but the covariant (flat-space)
derivative is actually zero.
The length Δα represents the "flat-space" derivative of
eα
with respect
to β in the α direction. The contribution of this change to the
derivative of ω is obtained by multiplying
that
length by ω
α, since that tells us how many
additional
lines of ω
eα will cross as a result of the change
-- and we can also read the number of new crossings directly from the
diagram (it's about 1).
The length Δγ represents the "flat-space" derivative of
eα
with respect
to β in the γ direction. The contribution of this change to
the derivative of ω is obtained by multiplying
that
length by ω
γ, since that tells us how many additional lines
of ω basis vector
eα crosses as a result this change
-- and we can also read the number of new crossings directly from the
diagram (it's about 2.7).
So, remembering that the
Christoffel symbols
measure how fast the basis vectors are changing relative to a
"flat-space" basis, the directional derivative we will actually measure
due to the parts
shown in the diagram will be (
no implied sum just yet!):
For spaces of higher dimension than 2, we can see that the α component
of the directional derivative in the direction β, for a
constant
1-form field, must be (
now we take an implied sum, over γ):
So, the actual covariant derivative must be the coordinate derivative,
minus
that value. When we sum across all components of a general vector
to
get the directional derivative with respect to that vector, we obtain:
which is the formula typically derived by non-visual (but more
rigorous) means in relativity texts.