Variation of scalar kinetic lagrangian

[OneLastTry]

Homework Statement

The goal of the question I'm being asked is to show that the covariant derivatives, D_{\mu}, "integrate by parts" in the same manner that the ordinary partial derivatives, \partial_{\mu} do.

More precisely, the covariant derivatives act on the complex scalar field \varphi_A such that
<br /> \begin{align}<br /> D_{\mu} \varphi_A &amp;= \partial_{\mu} \varphi_A - i g \varphi_A A^i_{\mu} T^{i \; B}_{\;A} \\<br /> D_{\mu} \varphi^{*A} &amp;= \partial_{\mu} \varphi^{*A} + ig \varphi^{*B} A^i_{\mu} T^{i \; A}_{\;B},<br /> \end{align}<br />
where T are the gauge group generators, g is the coupling constant and A is the gauge field.

I'm told to vary the scalar fields in the gauge-coupled scalar kinetic term, I = \int d^4 x \left[ -D^{\mu} \varphi^{*A} D_{\mu} \varphi_A \right] (subject to the usual fixing and fall-off conditions). Here's what the solutions say:

<br /> \begin{align}<br /> \delta I &amp;= \int d^4 x \left[ -\delta (D^{\mu} \varphi^{*A})D_{\mu} \varphi_A - D^{\mu} \varphi^{*A} \delta (D_{\mu} \varphi_A) \right] \\<br /> &amp;= \int d^4 x \left[ -(\partial_{\mu} \delta \varphi^{*A} + ig A^i_{\mu} T^{i \; A}_{\;B} \delta \varphi^{*B}) D^{\mu} \varphi_A - D^{\mu} \varphi^{*A} (\partial_{\mu} \delta \varphi_{A} - ig A^i_{\mu} T^{i\;B}_{\;A} \delta \varphi_B) \right]<br /> \end{align}<br />

From there, the professor integrates the first and third terms by parts (using the boundary conditions to set the surface term to zero), relabels and uses the definition of the covariant derivatives to show that you can integrate the covariant derivatives by parts in the same way as normal partial derivatives:

<br /> \delta I = \int d^4 x \left[ \delta \varphi^{*A} D_{\mu} D^{\mu} \varphi_A + \delta \varphi_A D_{\mu} D^{\mu} \varphi^{*A} \right].<br />

I understand all the parts except the bit that I'm sure someone will tell me is the most trivial :)
When varying the Lagrangian, why does
<br /> \delta (D^{\mu} \varphi^{*A})D_{\mu} \varphi_A = (\partial_{\mu} \delta \varphi^{*A} + ig A^i_{\mu} T^{i \; A}_{\;B} \delta \varphi^{*B}) D^{\mu} \varphi_A?<br />
The professor seems to have changed from upper to lower indices on the first part ( D^{\mu} \rightarrow \partial_{\mu} \dots + ig A^i_{\mu} \dots) and from lower to upper on the second part ( D_{\mu} \varphi_A \rightarrow D^{\mu} \varphi_A ).

I have a deep suspicion that I'm missing something obvious, but I can't see how those indices flipped. Can anyone help?

 

[member 553083]

Homework Equations The equations directly relevant to the question are:D_{\mu} \varphi_A = \partial_{\mu} \varphi_A - i g \varphi_A A^i_{\mu} T^{i \; B}_{\;A}D_{\mu} \varphi^{*A} = \partial_{\mu} \varphi^{*A} + ig \varphi^{*B} A^i_{\mu} T^{i \; A}_{\;B},I = \int d^4 x \left[ -D^{\mu} \varphi^{*A} D_{\mu} \varphi_A \right]\delta I = \int d^4 x \left[ -\delta (D^{\mu} \varphi^{*A})D_{\mu} \varphi_A - D^{\mu} \varphi^{*A} \delta (D_{\mu} \varphi_A) \right] The Attempt at a Solution I think that the professor is using the fact that raising and lowering indices with the metric tensor (in this case, the Minkowski tensor) doesn't change the components of the vector, so\partial_{\mu} = \eta_{\mu \nu} \partial^{\nu}and similarly for the gauge field.Is that correct?