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Comments to the question (v3): I) The notions of vectors, tensors, scalars, etc, depend on contexts in physics, cf. e.g. this and this Phys.SE posts and links therein. II) In OP's context, these notions refer to representations $\rho$ of the Lie group $SO(3)$ [and the corresponding Lie algebra $so(3)$] of 3D rotations, cf. e.g. Ref. 1. Let $\mathrm{i}L_k$, ...


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but I dont understand how $$\frac{1}{2}d(\vec v \cdot \vec v) = \frac{1}{2}\left(\vec v \cdot d\vec v + d\vec v \cdot v \right) = \vec v \cdot d\vec v$$ Also, I thought that $\vec{PM}\cdot d\vec{PM} = PMdPM$ if the angle between the two vectors is zero... While it's true that $\vec u \cdot \vec v = uv\cos\alpha$, it's also true that $$dv = ...


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The statement generalizes to all dimensions. Given a vector space $\mathbb{R}^n$ with the usual Euclidean metric, we represent a $k$-dimensional subplane spanned by the vectors $v_1,\dots,v_k$ with $v_1 \wedge v_2 \wedge \dots \wedge v_k$, where the wedge $\wedge$ is the antisymmetric linear product of the exterior algebra. Also, we have that the volume of ...


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This is a common misconception. When you apply a force upward on the object, the "reaction" force in Newton's 3rd Law is NOT the force of gravity down on the object; they do not have to be equal, and as you said, cannot be equal if you are to accelerate the object upwards. It is just a confusing coincidence that the force of gravity kind of looks like a ...


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The Poisson bracket you wrote only works for position (which is not a vector in general, as coordinates do not transform as vectors or any other tensor quantity). For general vectors the correct Poisson bracket is $$\{L_i,A_m\} = - (\mathbf p\times\nabla_{\mathbf p} + \mathbf r\times\nabla)_i A_m,$$ which reduces to the relation you wrote if you take $A_m = ...


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If $V$ is a (finite dimensional) vector space with no additional structure, then $V^*$ is a different vector space of the same dimension, hence isomorphic to $V$ --- but it would be a great mistake to think of $V$ and $V^*$ as "the same'' or even ``almost the same'', because there is no preferred isomorphism between them. In other words, you can pick a ...


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First, angular momentum isn't measured about an axis. It's measured about a point. Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in. Now, about ...


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Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, ...


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One way to prove it would be simply just do it the hard-way. Just find out the resolving vector. Lets assume there is a vector $\vec{C}$ which need to be resolved into vectors at direction $\alpha$ clockwise and $\beta$ counterclockwise to the Original vectors. Lets identify these directions by $\hat{A}_\alpha$ and $\hat{B}_\beta$ respectively. $$ i.e.\ \ ...


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Let $\mathbf v$ be any vector, and let $v$ denote its magnitude (i.e. $v^2=\mathbf v\cdot\mathbf v$). Then $$\mathbf v\cdot\text d\mathbf v=\frac12\text d(\mathbf v\cdot\mathbf v) = \frac12\text dv^2 = v\ \text dv.$$


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I dont understand how $\vec{PM}\cdot d\vec{PM} = \frac{1}{2}d(\vec{PM}\cdot \vec{PM})$ Read it from right to left and it's a special case of the product rule which holds for arbitrary bilinear functions including the scalar product, ie $$ d(\vec a\cdot\vec b) = \vec a\cdot d\vec b + d\vec a\cdot\vec b $$ The easiest way to see this is probably $$ (\vec ...



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