A question about the tensor of inertia components derivation

Question

Hello, I have gone through the derivation a few times, yet I still can’t grasp what is going on. First of all, how did the $xx^T$ term come to be? How can both $x^Tx$ and $xx^T$ be well defined at the same time considering $x$ is a vector? Also, I fail to see how that fourth equality is justified. If someone could please clarify the derivation for me, I will appreciate it.

classical mechanics moment of inertia

kaylimekay · Accepted Answer

The notation is confusing in the text you're looking at. There seems to be three different ways that the author is writing a vector dot product.
1: $mathbf a cdot  mathbf b$
2: $mathbf a^T mathbf b$
Both of these are the usual dot product $a_1 b_1 + a_2 b_2 + dots$. In the first case, the dot means to take the product in this way. In the second case, the superscript T means transpose, so $mathbf a^T$ is a row vector while $mathbf b$ is a column vector and you do matrix multiplication the usual way. The result is the same.
3: $mathbf a^2$
This notation is sloppy, but the author means you take the dot product of the vector with itself, i.e. compute its magnitude squared: $mathbf a^2 =mathbf a cdot mathbf a = mathbf a^T mathbf a$.
Actually there's one place where neither a dot nor a T is written but the dot product is implied. Keep in mind that having taken a dot product of two vectors, you obtain a scalar, i.e. just a "plain number." So after the 3rd equality in the first line you have an expression like $mathbf x (mathbf x cdot mathbf{hat n})mathbf{ hat n}$. The dot product $(mathbf x cdot mathbf{hat n})$ is just a scalar now so it can just be moved to the front of the expression with the 2. What's left is $mathbf x mathbf{hat n}$ by which the author sloppily means $mathbf x cdot mathbf{hat n}$. You therefore see that this whole term is equivalent to $-2(mathbf x cdot mathbf{hat n})^2$.
Now in the third term there, you have $+(mathbf x cdot mathbf{hat n})^2 mathbf{hat n}^2$. But $mathbf{hat n}$ is a unit vector so its magnitude squared is 1 by definition. So you see this term is the same as the second term except it comes with a factor of +1 instead of -2. Then $-2+1=-1$ so that explains the final equality.
Finally in the second line we see things like $mathbf x mathbf x^T$. This a matrix. For example $M=mathbf a mathbf b^T$ is a matrix whose components are given by products of the components of the vectors: $M_{ij}=a_i b_j$. There's no conflict with having both $mathbf x mathbf x^T$ and $mathbf x^T mathbf x$ at the same time. They just mean different things.
Note that dot product multiplication commutes so $mathbf xcdot mathbf{hat n} = mathbf{hat n}cdot mathbf x$. With that in mind you can write $(mathbf x cdot mathbf{hat n})^2 = (mathbf x cdot mathbf{hat n})(mathbf x cdot mathbf{hat n}) = (mathbf{hat n}cdot mathbf x)(mathbf x cdot mathbf{hat n})=(mathbf{hat n}^T mathbf x)(mathbf x^T mathbf{hat n})$, having switched between the various notational conventions ;).
But the parentheses don't really matter here, and you could just as well imagine forming the matrix $mathbf x mathbf x^T$ first, and then multiplying it on both sides by $mathbf{hat n}$ later, instead of taking the dot products first and multiplying the resulting scalars second.
The notation in this text sample was really bad, so don't feel discouraged.

Eli · Answer

$$vec u=(vec x^T,vec n),vec n$$
begin{align*}
   & text{$vec u~$ is also}
   &vec{u}=vec{n},vec{n}^T,vec x=
   left[ begin {array}{ccc} {n_{{x}}}^{2}&n_{{x}}n_{{y}}&n_{{x}}n_{{z}
} n_{{x}}n_{{y}}&{n_{{y}}}^{2}&n_{{y}}n_{{z}}
 n_{{x}}n_{{z}}&n_{{y}}n_{{z}}&{n_{{z}}}^{2}
end {array} right]vec x
=boldsymbol A,vec{x}
&text{with}
&boldsymbol A^T=A
&boldsymbol A^T,A=A,A=vec{n},vec{n}^T,vec{n},vec{n}^T=boldsymbol A
end{align*}
where the magnitude of $vec{n}^T,vec{n}=1$ .
and
$$vec R=vec x-vec u$$
$Rightarrow$
begin{align*}
  &vec{R}^T,vec{R}&=(vec{x}-vec u)^T,(vec{x}-vec u)&=
  vec{x}^T,vec{x}-2,vec{x}^T,vec{u}+vec{u}^T,vec{u}&=
   vec{x}^T,vec{x}-2,vec{x}^T,boldsymbol A,vec{x}+(boldsymbol Avec{x})^T,boldsymbol Avec{x}&=
   vec{x}^T,vec{x}-2,vec{x}^T,boldsymbol A,vec{x}+ vec{x}^T,boldsymbol Aboldsymbol Avec{x}&=
   vec{x}^T,vec{x}-2,vec{x}^T,boldsymbol A,vec{x}+ vec{x}^T,boldsymbol Avec{x}&=
   vec{x}^T,vec{x}- vec{x}^T,boldsymbol Avec{x}=
vec{x}^T,vec{x}- (vec{x}^T,vec{n})cdot ,(vec{n}^T,vec{x})&=
vec{x}^T,vec{x}- (vec{n}^T,vec{x})cdot ,(vec{x}^T,vec{n})
end{align*}
$$boxed{I=m,left(vec{R}^T,vec{R}right),E_3=m,left[vec{x}^T,vec{x}- (vec{n}^T,vec{x})cdot ,(vec{x}^T,vec{n})right],E_3}$$

John Alexiou · Answer

Another way of getting there is to say the distance $r$ is derived from $r = | boldsymbol{x} times boldsymbol{hat{n}}|$ provided that $|boldsymbol{hat{n}}|=1$. You have to use some vector triple product identities in the process, but it is all straight forward.
$$begin{aligned}I=mr^{2} & =mleft(boldsymbol{x}timesboldsymbol{hat{n}}right)cdotleft(boldsymbol{x}timesboldsymbol{hat{n}}right) &  & text{use }|boldsymbol{g}|=sqrt{boldsymbol{g}cdotboldsymbol{g}}
 & =mleft(boldsymbol{z}cdotleft(boldsymbol{x}timesboldsymbol{hat{n}}right)right) &  & text{assign }boldsymbol{z}text{ to make triple vector product}
 & =mleft(left(boldsymbol{x}timesboldsymbol{hat{n}}right)cdotboldsymbol{hat{n}}right) &  & text{triple product identity}
 & =mleft(left(left(boldsymbol{x}timesboldsymbol{hat{n}}right)timesboldsymbol{x}right)cdotboldsymbol{hat{n}}right) &  & text{expand }boldsymbol{z}
 & =mleft(left(boldsymbol{hat{n}}left(boldsymbol{x}cdotboldsymbol{x}right)-boldsymbol{x}left(boldsymbol{x}cdotboldsymbol{hat{n}}right)right)cdotboldsymbol{hat{n}}right) &  & text{vector triple product }
 & =mboldsymbol{hat{n}}^{top}left(left(boldsymbol{x}^{top}boldsymbol{x}right)mathrm{E}_{3}boldsymbol{hat{n}}-boldsymbol{x}left(boldsymbol{x}^{top}boldsymbol{hat{n}}right)right) &  & text{convert }(boldsymbol{a}cdotboldsymbol{b})text{ to }(boldsymbol{a}^{top}boldsymbol{b})
 & =mboldsymbol{hat{n}}^{top}left(left(boldsymbol{x}^{top}boldsymbol{x}right)mathrm{E}_{3}boldsymbol{hat{n}}-left(boldsymbol{x}boldsymbol{x}^{top}right)boldsymbol{hat{n}}right) &  & text{outer product identity }
end{aligned}
 $$
In my opinion, using cross products for mass moment of inertia is superior, especially when you understand the (deferred) cross product operator $[boldsymbol{x} times]$ representing the left-hand side of 3×3 matrix-vector product $boldsymbol{x} times boldsymbol{y} = [boldsymbol{x}times] boldsymbol{y}$. The definition is
$$ [pmatrix{x  y  z} times] = begin{bmatrix} 0 & -z & y  z & 0 & -x  -y & x & 0 end{bmatrix}$$
Then the mass moment of inertia 3×3 tensor is
$$ mathbf{I} =  m left( -[boldsymbol{x}times] [boldsymbol{x}times] right) $$
or by component
$$ mathbf{I} = m begin{bmatrix} y^2+z^2 & -x y & -x z  -x y & x^2+z^2 & -y z   -z x & -y z & x^2+y^2 end{bmatrix} $$
To prove this, consider a particle orbiting about the origin with rotational velocity $boldsymbol{omega}$. Its momentum is $boldsymbol{p} = m ( boldsymbol{omega} times boldsymbol{x})$. The angular momentum about the origin is
$$ boldsymbol{L} = boldsymbol{x} times boldsymbol{p} = m ( boldsymbol{x} times ( boldsymbol{omega} times boldsymbol{x} ) ) =  m ( - boldsymbol{x} times (boldsymbol{x} times boldsymbol{omega})) $$
From the above, you extract the 3×3 inertia matrix
$$ boldsymbol{L} = underbrace{ left( -m [boldsymbol{x} times] [boldsymbol{x} times] right) }_{mathbf{I}} boldsymbol{omega}$$

A question about the tensor of inertia components derivation

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