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I have question given as: $$\partial_k \varphi = -(C_k+ D_{jk}r_j)$$ where $C_k \,\&\, D_{jk}$ are constants and $D_{jk}$ is symmetric and traceless. I have to find $\varphi$.

I am getting : $\varphi = A -C_kr_k - D_{jk}r_jr_{k}$

but answer is: $\varphi = A -C_mr_m - \frac12 D_{sm}r_sr_{m}$ I am clueless about $\frac12$ term in the answer.

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I think your answer and the official answer are basically the same but they used the fact that the tensor D is symmetric and there is on additional little problem in your solution. Let me start from the beginning: \begin{equation} \partial_{k}\phi=-D_{jk}r_{j} \end{equation} this expression can be written as: \begin{equation} \partial_{k}\phi=-\frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}+D_{kj}r_{j}\right)- D_{kk}r_{k} \end{equation} Where I just used symmetry.This is easily integrated as: \begin{equation} A-\frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}r_{k}+D_{kj}r_{j}r_{k}\right)- \frac{1}{2}D_{kk}r_{k}r_{k} \end{equation} Now since any term that does not contain $r_{k}$ is just a constant in this moment you can and subtract this quantity to the expression: \begin{equation} \pm\sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m} \end{equation} where the plus term goes into A and the minus allows you the get the exact expression.

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  • $\begingroup$ Btw how do you cancel an extra term, $$ \pm\sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m} $$ I am not getting it, can you explain it? $\endgroup$
    – Zhang Wei
    Jul 23 at 10:20
  • $\begingroup$ The constant of integration is in general a function of $r_{i\ne k}$ and as shown by other methods this function must be a constant plus the negative branch of this term. $\endgroup$
    – Chris Long
    Jul 23 at 10:54
  • $\begingroup$ $$( A + \sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m}) - \frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}r_{k}+D_{kj}r_{j}r_{k} + \sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m} \right) $$ How this simplifies?@ChrisLong $\endgroup$
    – Zhang Wei
    Jul 23 at 11:28
  • $\begingroup$ Well I just meant that since A can be anything that is not a function of $r_{k}$ you can just say that: \begin{equation}A+\sum_{s\neq k \;\; m\neq k} D_{sm}r_{s}r_{m} \rightarrow A\end{equation} $\endgroup$
    – Yepman
    Jul 23 at 14:56
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Welcome to Physics StackExchange!

If you consider differentiating the third term:

$$\partial_i\left(-\frac{1}{2}D_{jk}r_jr_k\right)$$

First pull the constants put front:

$$-\frac{1}{2}D_{jk}\partial_i\left(r_jr_k\right)$$

Then apply product rule to $r_jr_k$ and see what you get. As this is a homework question I won't go further and want to let you try with this hint, but I am happy to add more if you still don't get it.

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    $\begingroup$ if I go from the answer to the question, I understand that $$\frac12 D_{sm}\partial_k r_s r_m =\frac12 D_{sm}(r_m \delta_{ks} + r_s\delta_{km}) = \frac12(D_{km}r_m + D_{ks}r_s) = D_{jk}r_j $$ $\endgroup$
    – Zhang Wei
    Jul 23 at 7:51
  • $\begingroup$ but how will I start from question only that : $$\int \partial r_k \ D_{jk}r_j$$ $\endgroup$
    – Zhang Wei
    Jul 23 at 7:53
  • $\begingroup$ Correct, so thats your factor of a half as the term is quadratic. $\endgroup$
    – Chris Long
    Jul 23 at 7:55
  • $\begingroup$ Well integration is the inverse of differentiating and is normally done by inspection. This is our inspection that we inverse to get the half. $\endgroup$
    – Chris Long
    Jul 23 at 7:56
  • $\begingroup$ indeed! Thanks for welcoming on the stack $\endgroup$
    – Zhang Wei
    Jul 23 at 8:00

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