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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)

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Hi, aries0152. I cannot download the paper at the moment, but are you sure with the index order? I am asking because there is $k$ three times in the expression. Should it be $\epsilon^{ijk}\partial_{i}B_{j}\dot{B}_{k}$ ? – Grisha Kirilin Jul 29 '12 at 15:10
Oh! My mistake!! The last term should be $\dot{B}$ not $\dot{B_k}$. The term was something like $2\mu^2(\nabla^2)^{-1}(\nabla \times B)\dot{B}$ So I just re-write the $\nabla \times B= \varepsilon^{ijk} \partial_j B_k \dot{B} $ – aries0152 Jul 29 '12 at 16:02
So it's a vector valued integral? Integrating those by parts is non-unique if you don't show your basis vectors or tell us some assumption about the basis of the system. – Jerry Schirmer Jul 29 '12 at 16:17
Still looks very strange. If $\dot{B}$ is the time derivative of $B_{k}$, then it should be a vector. I think the expression $(\nabla\times B) \dot{B}$ should be understood as a triple product, i.e., $(\epsilon^{ijk}\nabla_{i}B_{j}) \dot{B}_{k}$. – Grisha Kirilin Jul 29 '12 at 16:17
Actually this part should be zero, or somehow I have to make it zero. I have calculated the rest. Only this term causes the problem. One of my professor suggested me that try to use integration by parts. That might set it to zero! I have uploaded the Paper I substitute equation (24) into (22) but it does not satisfy the result of (25). I wish I could eliminate this term somehow! – aries0152 Jul 29 '12 at 17:27
up vote 1 down vote accepted

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 $I=0$.

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Thanks a lot. That was really a great help. :) – aries0152 Jul 30 '12 at 15:08

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