Multipole Expansion: Show that the quadrupole moment is symmetric and that the trace vanished

[Lambda96]

Homework Statement

Show that the quadrupole moment is symmetric and that the trace vanished

Relevant Equations

none

Hi

i have problems, to solve task a)

Since I have to calculate the trace of the matrix $Q$, I started as follows:

$$\text{trace} (Q)=\sum\limits_{i=1}^{3}\int_{}^{}d^3x'(3x_i^{'2}-r^{'2}) \rho(x')$$

I then calculated further until I got the following form:

$$\text{trace} (Q)=\int_{}^{}d^3x' \cdot 2x'^{2} \rho(x')$$

Now I'm stuck because I don't know how to show that the expression is 0

 

[TSny]

$$\text{trace} (Q)=\sum\limits_{i=1}^{3}\int_{}^{}d^3x'(3x_i^{'2}-r^{'2}) \rho(x')$$

I then calculated further until I got the following form:

$$\text{trace} (Q)=\int_{}^{}d^3x' \cdot 2x'^{2} \rho(x')$$

In getting to the second equation above, you made a mistake when evaluating $\sum\limits_{i=1}^{3}r^{'2}$.

 

[kuruman]

In getting to the second equation above, you made a mistake when evaluating $\sum\limits_{i=1}^{3}r^{'2}$.

I think the mistake is in dropping the subscript and writing $x'^2$ in OP's expression for the trace. What does that even mean?
(I deleted my post because it went in an unnecessary direction.)

 

[TSny]

I think the mistake is in dropping the subscript and writing $x'^2$ in OP's expression for the trace. What does that even mean?

I interpreted the OP's $x'^2$ to mean the same as $r'^2$.

I tried to guess how the OP got the incorrect factor of $2x'^2$. The sum over $i$ of $3{x'_i}^2$ gives $3r'^2$. Then I guessed that the OP just subtracted the $r'^2$ in the expression to get $2r'^2$ which they wrote as $2x'^2$. But, I shouldn't have done all this guessing! Bad form on my part.

Hopefully, the OP will clarify what they did.

 

[PhDeezNutz]

Hint: $(3x^2 - r^2 ) + (3y^2 - r^2) + (3z^2 - r^2)$ equals what?

 

[Lambda96]

Thank you TSny, kuruman and PhDeezNutz for your help

You are right TSny I proceeded as follows:

I interpreted $r'^2$ as follows:

$$r'^2=\sum\limits_{i=1}^{3}{x'}_i \cdot {x'}_i=\sum\limits_{i=1}^{3}{x'^2}_i={x'^2}_1+{x'^2}_2+{x'^2}_3$$

Then I calculated the following

$$\sum\limits_{i=1}^{3} \int_{}^{}d^3x'(3{x'^2}_i-{x'^2}_i) p(x’)$$
$$\sum\limits_{i=1}^{3} \int_{}^{}d^3x'\cdot 2{x'^2}_i p(x')$$
$$\int_{}^{}d^3x'\cdot 2({x'^2}_1+{x'^2}_2+{x'^2}_3) p(x')$$
$$\int_{}^{}d^3x'\cdot 2{x’^2} p(x')$$

In the end I should have written ${r’}^2$ instead of ${x'}^2$.

 

[PhDeezNutz]

I strongly suggest you write out the terms in the sum explicitly. There’s only 3.

The first term is

$3{x’}_1^2 - r^2$ where $r^2 = {x’}_1^2 + {x’}_2^2 + {x’}_3^2$

Write out the next two terms and sum all 3 terms together. What do you realize?

@Lambda96 post is fixed now.

I think I see what you are doing and while it is valid I don't think it's the best approach. Simplify at the end when you have written out all terms and summed them instead of trying to prove each term is zero ( I think that's what you're trying to do). Think about it this way; if the trace is $0$ regardless of charge distribution then the (entire) integrand (after you have summed) has to be categorically $0$.

Long and short of it: bring the sum inside the integral.

Don't prove the integral is zero by integrating each term.

Prove the integral is zero by proving the integrand is zero.

 

[Lambda96]

Thank you PhDeezNutz for your help

Do you mean the following?

$$=\sum\limits_{i=1}^{3} \int_{}^{}d^3x'(3{x'^2}_i-{r'}^2) p(x’)$$
$$=\sum\limits_{i=1}^{3} \int_{}^{}d^3x'(3{x'^2}_i-{x'^2}_1-{x'^2}_2-{x'^2}_3) p(x’)$$
$$=\int_{}^{}d^3x'(\sum\limits_{i=1}^{3} 3{x'^2}_i-{x'^2}_1-{x'^2}_2-{x'^2}_3) p(x’)$$
$$=\int_{}^{}d^3x'(3{x'^2}_1-{x'^2}_1-{x'^2}_2-{x'^2}_3+3{x'^2}_2-{x'^2}_1-{x'^2}_2-{x'^2}_3+3{x'^2}_3-{x'^2}_1-{x'^2}_2-{x'^2}_3) p(x’)$$
$$=\int_{}^{}d^3x'(3{x'^2}_1+3{x'^2}_2+3{x'^2}_3-3{x'^2}_1-3{x'^2}_2-3{x'^2}_3) p(x’)$$
$$=\int_{}^{}d^3x'0 p(x’)$$
$$=0$$

 

[kuruman]

More simply you change the order of summation and integration and then work on the summation to write $$\begin{align} & \sum\limits_{i=1}^{3} (3{x'^2}_i-{r'}^2)=\sum\limits_{i=1}^{3} (3{x'^2}_i)-\sum\limits_{i=1}^{3} ({r'}^2)\nonumber \\
& =3\sum\limits_{i=1}^{3} ({x'^2}_i)-({r'}^2)\sum\limits_{i=1}^{3}1=3({r'}^2)-({r'}^2)\times 3=0. \nonumber
\end{align}$$

 

[PhDeezNutz]

That’s exactly what I meant. Good Job.

 

[Lambda96]

Thank you kuruman and PhDeezNutz for your help