# Reason out the cross product (for the moment): a skew symmetric form

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• Trying2Learn

#### Trying2Learn

TL;DR Summary
Skew symmetric form of cross product
I am sure you are all familiar with the cross product in 3D space.

i cross into j gives k.
Cyclic
Negative, if reversed, etc.

I am sure you are all familiar with the definition as: norm of the first vector, norm of the second, sine of the angle, perpendicular (but direction using right hand rule)

Sure, no problem.

And we can apply to the Moment = Distance cross Force.

And I can reason my way through why the geometric rule with the sine is justified, and, if the space is Euclidean, how we get the cyclic rule, above, too!

All fine.

But there is another way, as (I am sure) you all know.

Take the first vector in the cross product and use the components to form a skew symmetric matrix with positive determinant.

OK...

Has anyone ever "reasoned" their way as to why this skew symmetric matrix, times the column components of the second vector can give the same information as the traditional cross product?

I attach a picture, because I am not good with the equation writer here.

You see, I can reason out the trig-based operation (sine of the distance, magnitudes, etc, but I cannot reason out why this works... I can only say it is isomorphic to the traditional way of i cross j gives k.

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• Cross.JPG
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But that video only speaks about the determinant. Unless I am confused as to the video (I could be), that does not mention (or reason) how the skew symmetric form of the first vector can be used to compute the cross product.

IT only mentions building BOTH vectors into ONE matrix, with the top row as the components: and I understand that.

I am looking for a "reasoning" as to why the skew symmetric form of the first vector can lead to the cross product. RIght now, I use my reasoning for the MOMENT of a force and the Trig rule for the cross product and then just say to myself: "the skew symmetric form implements THAT." And then I reason why THAT OLD Form gives the moment.

It's true, and arises because every vector is dual to an anti-symmetric tensor. Recall:\begin{align*}
(\mathbf{a} \times \mathbf{b})_i &= \epsilon_{ijk} a_j b_k \\
&\equiv A_{ik} b_k
\end{align*}with ##A_{ik} \equiv \epsilon_{ijk} a_j##. It's an antisymmetric matrix and its entries are ##A_{11} = 0##, then ##A_{12} = \epsilon_{132} a_3 = -a_3##, et cetera.

But that video only speaks about the determinant. Unless I am confused as to the video (I could be), that does not mention (or reason) how the skew symmetric form of the first vector can be used to compute the cross product.
You are right, in my memory there was more going on in that video... :)

It's true, and arises because every vector is dual to an anti-symmetric tensor. Recall:\begin{align*}
(\mathbf{a} \times \mathbf{b})_i &= \epsilon_{ijk} a_j b_k \\
&\equiv A_{ik} b_k
\end{align*}with ##A_{ik} \equiv \epsilon_{ijk} a_j##. It's an antisymmetric matrix and its entries are ##A_{11} = 0##, then ##A_{12} = \epsilon_{132} a_3 = -a_3##, et cetera.
That, too, I understand.

I am so sorry if I am not making this clear.

I am trying to find a way to "reason out" (perhaps with pictures), why I can use the skew symmetric form in the cross product.

In other words, If I want the moment of a force, I can say this: "The length of the first vector, times the length of the second, times the perpendicular moment arm (thus the sine function). Then, using Euclidean space, I can turn the operation into words like "i cross j is k."

But I am UNABLE to BEGIN with the skew symmetric form and reason out, with pictures, why that works to give a useful operation for the moment.

In terms of the exterior algebra ##\Lambda \mathbf{R}^3## over ##\mathbf{R}^3##, the cross product of two vectors ##\mathbf{a} \times \mathbf{b}## is the Hodge dual (##\star : \Lambda^p \mathbf{R}^3 \rightarrow \Lambda^{3-p} \mathbf{R}^3##) of the bivector ##\mathbf{a} \wedge \mathbf{b}##, i.e. ##\mathbf{a} \times \mathbf{b} = \star (\mathbf{a} \wedge \mathbf{b})##. The geometrical interpretation of ##\mathbf{a} \wedge \mathbf{b}## is a directed parallelogram!

In terms of the exterior algebra ##\Lambda \mathbf{R}^3## over ##\mathbf{R}^3##, the cross product of two vectors ##\mathbf{a} \times \mathbf{b}## is the Hodge dual (##\star : \Lambda^p \mathbf{R}^3 \rightarrow \Lambda^{3-p} \mathbf{R}^3##) of the bivector ##\mathbf{a} \wedge \mathbf{b}##, i.e. ##\mathbf{a} \times \mathbf{b} = \star (\mathbf{a} \wedge \mathbf{b})##. The geometrical interpretation of ##\mathbf{a} \wedge \mathbf{b}## is a directed parallelogram!
And that, too, I am aware of, but is not what I am looking for.

I wonder if what I am looking for is much simpler than I am explaining.

Here, I tried typing this picture. This is the best justification I can make.

It is a planar issue, but when I extend this reasoning to 3 dimensions I can "reason" my way as to why the skew symmetric form leads to an angular momentum.

But this is the best I can come up with. Has anyone reasoned out how "the skew symmetric form of the first vector's components, times the column of the second vector gives (this could be where my question confused everyone and I apologize for that -- previously I wrote "gives the cross product, but now I will write) "gives meaningful information."

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I'm still not sure what the question is. Do you mean why the angular momentum is what it is? It's because it's by definition the conserved quantity corresponding to symmetry under infinitesimal rotation ##\delta x^i = \epsilon_{ijk} (\delta \theta)_j x_k## where ##(\delta \theta)_j## is taken to be arbitrary. Under this symmetry transformation the Lagrangian varies like
\begin{align*}
\delta L = \frac{\partial L}{\partial x^i} \delta x^i + \frac{\partial L}{\partial \dot{x}^i} \delta \dot{x}^i &= \left(\frac{d}{dt} \frac{\partial L}{\partial \dot{x}^i} \right) \delta x^i + \frac{\partial L}{\partial \dot{x}^i} \delta \dot{x}^i = \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}^i} \delta x^i \right)
\end{align*}and putting ##\delta L = 0## means ##\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}^i} \delta x^i \right) = (\delta \theta)_j \frac{d}{dt} (\epsilon_{ijk} x_k p_i)##, or in other words this thing ##\epsilon_{ijk} x_j p_k \equiv L_i## (permuting the indices a bit) is constant. And we already discussed that in 3-dimensional space, ##L_i = \frac{1}{2} \epsilon_{ijk} L_{jk}## is dual to a two-index object ##L_{ij}##.

I'm still not sure what the question is. Do you mean why the angular momentum is what it is? It's because it's by definition the conserved quantity corresponding to symmetry under infinitesimal rotation ##\delta x^i = \epsilon_{ijk} (\delta \theta)_j x_k## where ##(\delta \theta)_j## is taken to be arbitrary. Under this symmetry transformation the Lagrangian varies like
\begin{align*}
\delta L = \frac{\partial L}{\partial x^i} \delta x^i + \frac{\partial L}{\partial \dot{x}^i} \delta \dot{x}^i &= \left(\frac{d}{dt} \frac{\partial L}{\partial \dot{x}^i} \right) \delta x^i + \frac{\partial L}{\partial \dot{x}^i} \delta \dot{x}^i = \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}^i} \delta x^i \right)
\end{align*}and putting ##\delta L = 0## means ##\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}^i} \delta x^i \right) = (\delta \theta)_j \frac{d}{dt} (\epsilon_{ijk} x_k p_i)##, or in other words this thing ##\epsilon_{ijk} x_j p_k \equiv L_i## (permuting the indices a bit) is constant. And we already discussed that in 3-dimensional space, ##L_i = \frac{1}{2} \epsilon_{ijk} L_{jk}## is dual to a two-index object ##L_{ij}##.
Foremost: thank you for your continued patience as I try to explain my issue...

Let me try this...

There is a need, in say, statics, (does not have to be angular momentum) to keep track of this information:
1. the force and its magnitude
2. the distance to the force from some origin (its magnitude)
3. The angle between the distance and the force
The trig/geometric definition (Let me call this: IMPLEMENTATION 1) of the cross product satisfies this need:
• norm of the first
• norm of the second
• sine of the angle between and
• the plane in which they act (the perpendicular vector)

From that, in Euclidean space, I can derive the other algebraic implementation:
i cross j = k (cyclic, right hand rule, etc.)
But let me call that: IMPLEMENTATION 2

Now I do not like IMPLEMENTATION 2. The use of the skew symmetric form of the first vector, times the column of the second, is easier to code. Let me call this: IMPLEMENTATION 3

However, I must then make an isomorphic map between IMPLEMENTATION 2 and 3.

I would prefer to go DIRECTLY to implementation 3 without the reasoning used to get implementation 1.

That is the best way I can say it

So, if you do part of the matrix manipulation you just get:
$$\bar v \times \bar w = [\hat i \ \hat j \ \hat k] \begin{bmatrix} w_3 v_2 -w_2 v_3 \\ w_1 v_3 - v_1 w_3 \\ w_2 v_1 - w_1 v_2 \end{bmatrix}$$
Are you simply asking why the terms in the right matrix look the way they do? Because these are just the terms for a cross product, which you will only get by using the skew symmetric matrix. The closest I can come for this is showing you this video (which is the one I wanted to link to in my first post):

So, if you do part of the matrix manipulation you just get:
$$\bar v \times \bar w = [\hat i \ \hat j \ \hat k] \begin{bmatrix} w_3 v_2 -w_2 v_3 \\ w_1 v_3 - v_1 w_3 \\ w_2 v_1 - w_1 v_2 \end{bmatrix}$$
Are you simply asking why the terms in the right matrix look the way they do? Because these are just the terms for a cross product, which you will only get by using the skew symmetric matrix. The closest I can come for this is showing you this video (which is the one I wanted to link to in my first post):

Close... really close

But you wrote this: "Because these are just the terms for a cross product, which you will only get by using the skew symmetric matrix"

My issue is this. I want to reason out that form WITHOUT referring to the traditional terms of the cross product... And I want to reason it for Moments and Angular momentum.

And again: THANK YOU for your ongoing patience. Half my problem is my inability to ask the question.

In other words: look at my simple attempt in the second picture I posted. Maybe that is really all I can say.

I would prefer to go DIRECTLY to implementation 3 without the reasoning used to get implementation 1.
I'm not sure what you are referring to. Deriving the angular momentum? A rotation is a matrix ##R_{ij}##, and more than that it's an orthogonal matrix ##R^T R = I## since it must preserve lengths:\begin{align*}
\mathbf{x}'^T \mathbf{x}' = (R \mathbf{x})^T R \mathbf{x} = \mathbf{x}^T R^T R \mathbf{x} \equiv \mathbf{x}^T \mathbf{x}
\end{align*}Consider an infinitesimal rotation with ##R_{ij} = \delta_{ij} + r_{ij}##, where the ##r_{ij}## is "small". The requrement of length-preservation means, to first order, that
\begin{align*}
(\delta^T_{ij} + r^T_{ij})(\delta_{jk} + r_{jk}) &= \delta_{ik} \\
(\delta_{ji} + r_{ji})(\delta_{jk} + r_{jk}) &= \delta_{ik} \\
r_{ki} + r_{ik} &= 0
\end{align*}i.e. that ##r_{ij}## is antisymmetric. Consider again the symmetry transformation ##x_i '= R_{ij} x_j = x_i + r_{ij} x_j##, or equivalently ##\delta x_i = r_{ij} x_j##, then from the same variational argument as in post #9 we have \begin{align*}
\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}_i} \delta x_i \right)&= r_{ij} \frac{d}{dt} (p_i x_j) \\
&= \frac{1}{2} r_{ij} \frac{d}{dt} (p_i x_j - p_j x_i)
\end{align*}due to the anti-symmetry of ##r_{ij}##, and finally you invoke the arbitrariness of ##r_{ij}## to conclude that the two-index object ##x_i p_j - x_j p_j## is conserved.

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I'm not sure what you are referring to. Deriving the angular momentum? A rotation is a matrix ##R_{ij}##, and more than that it's an orthogonal matrix ##R^T R = I## since it must preserve lengths:\begin{align*}
\mathbf{x}'^T \mathbf{x}' = (R \mathbf{x})^T R \mathbf{x} = \mathbf{x}^T R^T R \mathbf{x} \equiv \mathbf{x}^T \mathbf{x}
\end{align*}Consider an infinitesimal rotation with ##R_{ij} = \delta_{ij} + r_{ij}##, where the ##r_{ij}## is "small". The requrement of length-preservation means, to first order, that
\begin{align*}
(\delta^T_{ij} + r^T_{ij})(\delta_{jk} + r_{jk}) &= \delta_{ik} \\
(\delta_{ji} + r_{ji})(\delta_{jk} + r_{jk}) &= \delta_{ik} \\
r_{ki} + r_{ik} &= 0
\end{align*}i.e. that ##r_{ij}## is antisymmetric. Consider again the symmetry transformation ##x_i = R_{ij} x_j = x_i + r_{ij} x_j##, or equivalently ##\delta x_i = r_{ij} x_j##, then from the same variational argument as in post #9 we have \begin{align*}
\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}_i} \delta x_i \right)&= r_{ij} \frac{d}{dt} (p_i x_j) \\
&= \frac{1}{2} r_{ij} \frac{d}{dt} (p_i x_j - p_j x_i)
\end{align*}due to the anti-symmetry of ##r_{ij}##, and finally you invoke the arbitrariness of ##r_{ij}## to conclude that the two-index object ##x_i p_j - x_j p_j## is conserved.

However, let's say you are teaching undergraduate students, STATICS.

You arrive at the need to discuss the MOMENT of a force.

You apply the trig formula rule and reason your way that it is useful.

Now make the same argument, to undergraduates, on why using the skew symmetric form, is useful.

That is all I am looking for: a way to abandon the old cross product and its "i cross j, is k" rule

So, if I assume that you understand the relation between the cross product and the determinant, than what you are asking is the 3D equivalent of this? That's hard :) • suremarc