Coupling of vector gauge and a massive tensor field I was reviewing the paper-Coupling of a vector gauge field to a massive tensor field
In the calculation I found the term $ 2\mu^2 \varepsilon^{ijk} \dfrac{\partial_j}{\partial^2}B_k\dot{B}$ which needed to be integrated by parts. How can I do this? Please help me.
Note: Here $B_k$ is the space component of an antisymmetric tensor field and $\dot{B}$ is the time derivative of $B_k$.
(Edit: corrected the index)
 A: Actually the terms you interesting in is vanishing after integration over full
space-time domain. It is easy to see if you use Fourier transform:
$$
\mathbf{B}_{t,\mathbf{x}}=\int\frac{d^{4}k}{\left(  2\pi\right)  ^{4}
}\,\mathbf{B}_{\omega,\mathbf{k}}\,e^{-i(\omega t-\mathbf{k}\cdot\mathbf{x})}.
$$
such that
$$
\mathbf{B}_{\omega,\mathbf{k}}^{\star}=\mathbf{B}_{-\omega,-\mathbf{k}}
$$
Thus your term takes the form:
\begin{align*}
I  & =\int dt\int d^{3}x\,\mathbf{\dot{B}}\cdot\left(  \nabla^{2}\right)
^{-1}\left(  \mathbf{\nabla}\times\mathbf{B}\right)  =\int d^{4}
k\,\frac{\omega}{\mathbf{k}^{2}}\,\mathbf{B}_{-\omega,-\mathbf{k}}
\cdot\left[  \mathbf{k}\times\mathbf{B}_{\omega,\mathbf{k}}\right]  =\\
& =\int d^{4}k\,\frac{\omega}{\mathbf{k}^{2}}\,\mathbf{B}_{\omega,\mathbf{k}
}^{\star}\cdot\left[  \mathbf{k}\times\mathbf{B}_{\omega,\mathbf{k}}\right]  .
\end{align*}
Since $I$ is real, i.e., $I^{\star}=I$, we find:
$$
I=I^{\star}=\int d^{4}k\,\frac{\omega}{\mathbf{k}^{2}}\,\mathbf{B}
_{\omega,\mathbf{k}}\cdot\left[  \mathbf{k}\times\mathbf{B}_{\omega
,\mathbf{k}}^{\star}\right]  =-\int d^{4}k\,\frac{\omega}{\mathbf{k}^{2}
}\,\mathbf{B}_{\omega,\mathbf{k}}^{\star}\cdot\left[  \mathbf{k}
\times\mathbf{B}_{\omega,\mathbf{k}}\right]  =-I,
$$
hence $I=0$.
It is also easy to show that $I=0$ without using Fourier transform. As the first step, one can utilize the following property:
$$
\int d^{3}x\,A\left(  \nabla^{2}\right)  ^{-1}B=\int d^{3}x\,B\left(
\nabla^{2}\right)  ^{-1}A,
$$
thus
$$
I=\int dt\int d^{3}x\,\mathbf{\dot{B}}\cdot\left(  \nabla^{2}\right)
^{-1}\left(  \mathbf{\nabla}\times\mathbf{B}\right)  =\int dt\int
d^{3}x\left(  \mathbf{\nabla}\times\mathbf{B}\right)  \cdot\left[  \left(
\nabla^{2}\right)  ^{-1}\mathbf{\dot{B}}\right]  .
$$
Secondly, for the sake of convenience, one can use component notations:
$$
I=\epsilon^{ijk}\int dt\int d^{3}x\nabla_{i}B_{j}\left(  \nabla^{2}\right)
^{-1}\dot{B}_{k}=\epsilon^{ijk}\int dt\int d^{3}xB_{i}\left(  \nabla
^{2}\right)  ^{-1}\nabla_{j}\dot{B}_{k},
$$
where I use integration by parts (IBP) with respect to spatial dimensions and change
the order of indices in $\epsilon^{ijk}$. Then I would like to use IBP with
respect to time:
$$
I=-\epsilon^{ijk}\int dt\int d^{3}x\dot{B}_{i}\left(  \nabla^{2}\right)
^{-1}\nabla_{j}B_{k}=-I,
$$
hence $I=0$.
