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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.

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  • $\begingroup$ $x^T x$ is an inner product forming a scalar, while $xx^T$ is an outer product, forming a matrix. Which is the fourth equality? $\endgroup$ Dec 16, 2020 at 7:34

3 Answers 3

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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 $\mathbf x\cdot \mathbf{\hat n} = \mathbf{\hat n}\cdot \mathbf x$. 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.

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$$\vec u=(\vec x^T\,\vec n)\,\vec n$$ \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 $$\vec R=\vec x-\vec u$$

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\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 A\vec{x})^T\,\boldsymbol A\vec{x}\\&= \vec{x}^T\,\vec{x}-2\,\vec{x}^T\,\boldsymbol A\,\vec{x}+ \vec{x}^T\,\boldsymbol A\boldsymbol A\vec{x}\\&= \vec{x}^T\,\vec{x}-2\,\vec{x}^T\,\boldsymbol A\,\vec{x}+ \vec{x}^T\,\boldsymbol A\vec{x}\\&= \vec{x}^T\,\vec{x}- \vec{x}^T\,\boldsymbol A\vec{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*}

$$\boxed{I=m\,\left(\vec{R}^T\,\vec{R}\right)\,E_3=m\,\left[\vec{x}^T\,\vec{x}- (\vec{n}^T\,\vec{x})\cdot \,(\vec{x}^T\,\vec{n})\right]\,E_3}$$

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  • $\begingroup$ The factor $m$ is missing from the last equation. $\endgroup$ Dec 16, 2020 at 17:53
  • $\begingroup$ thank you i will correct it $\endgroup$
    – Eli
    Dec 16, 2020 at 18:14
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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.

$$\begin{aligned}I=mr^{2} & =m\left(\boldsymbol{x}\times\boldsymbol{\hat{n}}\right)\cdot\left(\boldsymbol{x}\times\boldsymbol{\hat{n}}\right) & & \text{use }|\boldsymbol{g}|=\sqrt{\boldsymbol{g}\cdot\boldsymbol{g}}\\ & =m\left(\boldsymbol{z}\cdot\left(\boldsymbol{x}\times\boldsymbol{\hat{n}}\right)\right) & & \text{assign }\boldsymbol{z}\text{ to make triple vector product}\\ & =m\left(\left(\boldsymbol{x}\times\boldsymbol{\hat{n}}\right)\cdot\boldsymbol{\hat{n}}\right) & & \text{triple product identity}\\ & =m\left(\left(\left(\boldsymbol{x}\times\boldsymbol{\hat{n}}\right)\times\boldsymbol{x}\right)\cdot\boldsymbol{\hat{n}}\right) & & \text{expand }\boldsymbol{z}\\ & =m\left(\left(\boldsymbol{\hat{n}}\left(\boldsymbol{x}\cdot\boldsymbol{x}\right)-\boldsymbol{x}\left(\boldsymbol{x}\cdot\boldsymbol{\hat{n}}\right)\right)\cdot\boldsymbol{\hat{n}}\right) & & \text{vector triple product }\\ & =m\boldsymbol{\hat{n}}^{\top}\left(\left(\boldsymbol{x}^{\top}\boldsymbol{x}\right)\mathrm{E}_{3}\boldsymbol{\hat{n}}-\boldsymbol{x}\left(\boldsymbol{x}^{\top}\boldsymbol{\hat{n}}\right)\right) & & \text{convert }(\boldsymbol{a}\cdot\boldsymbol{b})\text{ to }(\boldsymbol{a}^{\top}\boldsymbol{b})\\ & =m\boldsymbol{\hat{n}}^{\top}\left(\left(\boldsymbol{x}^{\top}\boldsymbol{x}\right)\mathrm{E}_{3}\boldsymbol{\hat{n}}-\left(\boldsymbol{x}\boldsymbol{x}^{\top}\right)\boldsymbol{\hat{n}}\right) & & \text{outer product identity } \end{aligned} $$

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

$$ \boldsymbol{L} = \underbrace{ \left( -m [\boldsymbol{x} \times] [\boldsymbol{x} \times] \right) }_{\mathbf{I}} \boldsymbol{\omega}$$

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