I Derivation of an identity for ##\partial^2_t \int T^{00}(x^i x_i)^2d^3

[MathematicalPhysicist]

TL;DR Summary

I want to show the identity:
$\partial^2_t \int T^{00}(x^i x_i)^2 d^3x = 4\int T^i_i x^j x_j d^3x + 8\int T^{ij}x_i x_j d^3x$
by using the identity: $T^{\mu \nu}_{,\nu}=0$, for a bounded system (i.e. a system for which $T^{\mu \nu}=0$ outside a bounded region of space).

I'll write down my calculations, and I would like if someone can point me to my mistakes.

$$\partial_t \int T^{00}(x^i x_i)^2 d^3 x = -\int T^{0k}_{,k}(x^i x_i)^2 d^3 x = \Dcancelto[0]{-\int (T^{0k}(x_ix^i)^2)_{,k}d^3 x} +\int (T^{0k}(x_i x^i)^2_{,k})d^3 x$$

After that:
$$\partial_t \int T^{0k} [ 2x_k (x^k)^2+2(x_k)^2 x^k] d^3 x = \int T^{0k},0 [ 2x_k (x^k)^2+2(x_k)^2 x^k] d^3 x = $$
$$=-\int T^{kj}_{,j}[2x_k(x^k)^2+2(x_k)^2 x^k] =\[ \Dcancelto[0]{-\int (T^{kj}[ 2x_k(x^k)^2+2(x_k)^2x^k])_{,j}d^3 x}\] + \int T^{kj}[2\delta_{jk}(x^k)^2+4x_k x^k \delta^k_j + 2(x_k)^2\delta^k_j+4x_k \delta_{jk} x^k ] d^3 x$$

How to continue? assuming my above work is correct?

Thanks!
The \Dcancelto should be the arrow that goes to zero, I am not sure how to fix this here.

 

[kent davidge]

The arrow is given by Longrightarrow.

 

[samalkhaiat]

someone can point me to my mistakes.
$$\partial_t \int T^{00}(x^i x_i)^2 d^3 x = \int T^{0k} \left( (x_i x^i)^2 \right)_{,k}\ d^3 x$$

\left( ( x_{i}x^{i} )^{2} \right)_{,k} = 4 x_{k} \left( x_{j}x^{j} \right) .

 

[MathematicalPhysicist]

\left( ( x_{i}x^{i} )^{2} \right)_{,k} = 4 x_{k} \left( x_{j}x^{j} \right) .

Isn't this should be: $(x_i x^i)^2_{,k} = (x_i x^i x_i x^i)_{,k}=2x_k (x^k)^2 + 2x^k (x_k)^2$?, where I used the fact that: $\delta_{ik} = x_{i,k}$.

 

[MathematicalPhysicist]

OK, I think I see my mistake (not really mine, it's the awful physicists' notation).
$(x_i x^i)^2$ is actually $(x_i x^i) (x_j x^j)$.

 

[Ibix]

$$\begin{eqnarray*}
&&((x_ix^i)^2)_{,k}\\
&=&(x_ix^ix_jx^j)_{,k}\\
&=&(x_ix^i)_{,k}(x_jx^j)+(x_ix^i)(x_jx^j)_{,k}\\
&=&2x_k(x_jx^j)+(x_ix^i)2x_k\\
&=&4x_k(x_jx^j)\end{eqnarray*}$$
...and I typed all that out so I'm posting it even though I see you worked it out before I'd quite finished. In defence of the notation, I don't think $(x_ix^i)^2$ is at all ambiguous. Expanding it as $(x_ix^ix_ix^i)$ breaks the "no index more than twice" rule. And it's not at all clear to me what your $x_k(x^k)^2$ could mean, because you are using $k$ as both a dummy and a free subscript, which should hint to you that you've done something wrong.

 

[MathematicalPhysicist]

$$\begin{eqnarray*}
&&((x_ix^i)^2)_{,k}\\
&=&(x_ix^ix_jx^j)_{,k}\\
&=&(x_ix^i)_{,k}(x_jx^j)+(x_ix^i)(x_jx^j)_{,k}\\
&=&2x_k(x_jx^j)+(x_ix^i)2x_k\\
&=&4x_k(x_jx^j)\end{eqnarray*}$$
...and I typed all that out so I'm posting it even though I see you worked it out before I'd quite finished. In defence of the notation, I don't think $(x_ix^i)^2$ is at all ambiguous. Expanding it as $(x_ix^ix_ix^i)$ breaks the "no index more than twice" rule. And it's not at all clear to me what your $x_k(x^k)^2$ could mean, because you are using $k$ as both a dummy and a free subscript, which should hint to you that you've done something wrong.

Yes, you are correct in saying that it's not ambiguous while remembering to use Einstein summation convention, my problem is that I was also been reading books in pure maths and they don't always use the Einstein summation convention, so it can be quite confusing going from one book to another.

Thanks anyway!