# Integration of tensor to find potential

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.

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: $$$$\partial_{k}\phi=-D_{jk}r_{j}$$$$ this expression can be written as: $$$$\partial_{k}\phi=-\frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}+D_{kj}r_{j}\right)- D_{kk}r_{k}$$$$ Where I just used symmetry.This is easily integrated as: $$$$A-\frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}r_{k}+D_{kj}r_{j}r_{k}\right)- \frac{1}{2}D_{kk}r_{k}r_{k}$$$$ Now since any term that does not contain $$r_{k}$$ is just a constant in this moment you can and subtract this quantity to the expression: $$$$\pm\sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m}$$$$ where the plus term goes into A and the minus allows you the get the exact expression.

• 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? Jul 23 at 10:20
• 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. Jul 23 at 10:54
• $$( A + \sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m}) - \frac{1}{2}\sum_{j\neq k}\left(D_{jk}r_{j}r_{k}+D_{kj}r_{j}r_{k} + \sum_{s\neq k \;\;m\neq k}D_{sm}r_{s}r_{m} \right)$$ How this simplifies?@ChrisLong Jul 23 at 11:28
• Well I just meant that since A can be anything that is not a function of $r_{k}$ you can just say that: $$A+\sum_{s\neq k \;\; m\neq k} D_{sm}r_{s}r_{m} \rightarrow A$$ Jul 23 at 14:56

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.

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