Dipole-Coupling Tensor: Electrostatic Dipole Moments

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I've been struggling with this problem today.

Here's an image of the question I'm attempting to answer.

I'm relatively new to tensor algebra (I've been studying it for about a week or two), and I've been able to solve most problems, except for this one. Here's some work:

$$ U = \frac{1}{r^3}\left(\vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2}\right) \equiv \frac{1}{r^3}P_iT_j^iQ^j\\ \vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2} = P_iT_j^iQ^j\\ $$ $$ P^ie_iQ^je_j- \dfrac{3(P^ie_i(Q^je_j - P^ie_i))(Q^je_j(Q^je_j - P^ie_i))}{(Q^je_j - P^ie_i)^2} = P_iT_j^iQ^j\\ $$ $$ P^ie_iQ^je_j- \dfrac{3P^ie_iQ^je_j(Q^je_j - P^ie_i)(Q^je_j - P^ie_i)}{(Q^je_j - P^ie_i)^2} = P_iT_j^iQ^j\\ P^ie_iQ^je_j- \dfrac{3P^ie_iQ^je_j(Q^je_j - P^ie_i)^2}{(Q^je_j - P^ie_i)^2} = P_iT_j^iQ^j\\ P^ie_iQ^je_j- 3P^ie_iQ^je_j = P_iT_j^iQ^j\\ -2P^ie_iQ^je_j = P_iT_j^iQ^j\\ -2(\delta^{ik}P_k)e_iQ^je_j = P_iT_j^iQ^j\\ P_k(-2\delta^{ik})e_ie_jQ^j = P_iT_j^iQ^j\\ P_k(-2\delta^{ik}\delta_{ij})Q^j = P_iT_j^iQ^j\\ P_k(-2\delta^{k}_{j})Q^j = P_iT_j^iQ^j\\ P_i(-2\delta^{i}_{j})Q^j = P_iT_j^iQ^j $$ $$ \boxed{T^i_j = -2\delta^i_j} $$

I'm not very confident that this is the correct answer. I feel as though I'm manipulating these tensors half-assed. Moreover, this answer does not appear to be very elegant; I wouldn't expect a coupling tensor to have this form (although perhaps if we replaced the kronecker delta with a metric tensor for curved space, I would be more compelled to believe it). In the last step, I changed indices because I believe you can do so with impunity.

Please let me know if you see any blatant errors that should not have been committed under any circumstances. I have found that doing independent studies with tensors has been the most difficult for me yet.

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So with the help of @Winther, the solution is as follows:

$$ U = \frac{1}{r^3}\left(\vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2}\right) \equiv \frac{1}{r^3}P_iT_j^iQ^j\\ \vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2} = P_iT_j^iQ^j\\ $$ $$ P^ke_kQ_le^l- \dfrac{3(P^ie_ir_je^j)(Q^ne_nr_me^m)}{r^2} = P_iT_j^iQ^j\\ P^lQ_l- \dfrac{3(P^ir_i)(Q^nr_n)}{r^2} = P_iT_j^iQ^j\\ P^lQ_l- \dfrac{3(\delta^{iv}P_vr_i)(Q^nr_n)}{r^2} = P_iT_j^iQ^j\\ P^lQ_l- \dfrac{3P_vQ^n(r^vr_n)}{r^2} = P_iT_j^iQ^j\\ P_aQ^a- \dfrac{3P_aQ^n(r^ar_n)}{r^2} = P_iT_j^iQ^j\\ P_a\delta^a_nQ^n- \dfrac{3P_a(r^ar_n)Q^n}{r^2} = P_iT_j^iQ^j\\ P_a\left(\delta^a_n - 3\dfrac{(r^ar_n)}{r^2}\right)Q^n = P_iT_j^iQ^j\\ P_a\left[T_n^a\right]Q^n = P_iT_j^iQ^j\\ $$ $$ \boxed{T_j^i = \delta^i_j - 3\dfrac{r^ir_j}{r^2}} $$