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Is it possible to write $$\int d^3x \,\,\, x_i\,\,x_j\,\,\,f(\vec x)$$ where $f(\vec x)$ is some function of the position and the indices indicate which component,

as a sum of a traceless tensor and a multiple of $\delta_{ij}$? If so, is there a good way of seeing how that may be achieved?

Thank you.

I don't think the following point is relevant, but just in case, I should add that $f(\vec x)$ decays as $|\vec x|\to \infty$

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up vote 1 down vote accepted

Define $$ M_{ij} = \int d^3x \, x_ix_jf(\vec x).$$

To make $M$ traceless, we subtract the appropriate multiple of identity, $$N_{ij} = M_{ij} - \frac{\mathop{\rm tr M}}3 \delta_{ij}.$$

It is easy to check that $\mathop{\rm tr N}=0$ and $$M_{ij} = N_{ij} + \frac{\mathop{\rm tr M}}3 \delta_{ij}$$ by definition.

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I take it $f$ is a scalar field. Then what you have is a linear operator $\underline T$ such that

$$x_i x_j f = \underline T(e_i;x) \cdot e_j \implies \underline T(a;x) = (a \cdot x) x f(x)$$

where $\underline T$ is linear in $a$ (but not $x$).

This operator is symmetric. See that the adjoint is

$$\overline T(b) = (b \cdot x) x f(x) = \underline T(b)$$

Can you make this into the sum of a traceless operator and the identity? Well, sure. Let's find the trace, $T$.

$$T = \nabla_a \cdot \underline T(a) = \nabla_a \cdot (a \cdot x) x f(x) = x^2 f(x)$$

Now you can construct a tracefree operator $\underline F$ by subtracting out $Ta/3$.

$$\underline F(a) = \underline T(a) - Ta/3 = (a \cdot x) x f(x) - ax^2 f(x)/3$$

The part with nonzero trace is the part we subtracted out, $Ta/3 = T \underline I(a)/3$. Remember, the Kronecker delta, $\delta_{ij}$, merely represents the components of the identity operator.

I leave it to you to evaluate the components $F_{ij}$ of the operator $\underline F$ now. In addition, while it was possible to make the integrand into a tensor field, making the whole integral into a non-field tensor is harder to do without some information about the integral itself.

At any rate, the key you should take away here is that you can always construct a tracefree linear operator just by subtracting out the identity operator multiplied by the trace.

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