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The dot product of two vectors $\textbf{A}$ and $\textbf{B}$ is given by $$\textbf{A}.\textbf{B}=ABcos\theta$$ where $A$ and $B$ are the magnitudes of the vectors $\textbf{A}$ and $\textbf{B}$, respectively, and $\theta$ is the angle between the vectors $\textbf{A}$ and $\textbf{B}$. In short, the dot product of $\textbf{A}$ with $\textbf{B}$ picks ...


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Take your first relation for the 3 Pauli matrices individually: $$\sigma_1(t)=U^\dagger\sigma_1(0)U$$ $$\sigma_1(t)=U^\dagger\sigma_2(0)U$$ $$\sigma_3(t)=U^\dagger\sigma_3(0)U$$ Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ...


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Gravity doesn't have a horizontal component. The component of gravity normal to the plane in your diagram can be said to have a horizontal component, sure (and a vertical component of magitude $mg\cos^{2}\theta$). But there is also a component of gravity parallel to the plane of magnitude $mg\sin{\theta}$. That component can be resolved into a vertical and ...


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There is a little bit more thinking behind saying that $P=\vec F \cdot \vec v$ than it being a generalised multiplication in 3D. There are even cases where multiplication with scalar becomes a cross product when using 3D vectors. For example, torque $T=Fr$, becomes $\vec T = \vec r \times \vec F$. Whenever implementing vectors into existing scalar equations, ...


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The scalar product is just a shortcut notation for multiplication and then addition: $$\vec{a}\cdot\vec{b} = a_x b_x + a_y b_y + a_z b_z\tag{1}$$ It's commutative if the underlying multiplication is commutative, and otherwise it is not. Notation like $\vec{a}\cdot\nabla$ is not really a scalar product, but it takes the same form of (1) and applies it to ...


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No. Trivial counterexample would be if the origin, o, is just below the tangent point for v1, directly below it. In that case, beta clearly can be 90 degrees or more, even if v1 and v2 are actually in the same direction!


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These two formulae must give the same result. I think there are some mistakes in your calculation. The result a) is correct. a) $(ia_{\mu} + b_{\mu}) = i(a_{\mu} - ib_{\mu} )$ and then it is $$ i^2(a_{\mu}- ib_{\mu} )(a^{\mu}+ ib^{\mu} ) = i^2(a^2 + ia_{\mu}b^{\mu}-ia^{\mu}b_{\mu} - ib^{2}) = i^2(a^2+b^2) $$ . But result b) is wrong. The correct result is $$ ...


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A vector is a geometric object with both direction and "length." Magnitude is basically the length of the vector from head to tail. The magnitude therefore has to be positive since lengths have to be positive. A component is any individual entry in the vector. For example, the vector $\vec{v}=\begin{pmatrix} 228 \\ 43 \\ 392 \\ 25 ...


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To add to Dilithium Matrix's answer, to simplify, you can look at $v_1$ and $v_2$ as any vectors, not necessarily velocity vectors. We know that if $a = b$ and $b = c$, then $a = c$. Hence, to say that $\theta = \beta$ is to say that $\alpha = \beta$. You can then use geometry to show that there is only one other case where, for $O' \neq O$, $\alpha$ will ...


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This can be a little subtle the first time you see it, so I'll move through the rationale carefully: $\langle (\delta \mathbf{r} \cdot \nabla )^2 \rangle _{vac} = \langle (\delta x \ \partial_x + \delta y\ \partial_y + \delta {z}\ \partial_z )^2\rangle _{vac} $ On expanding you get squared and cross terms. The text mentions $\langle \delta \mathbf r ...


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Recall that the dot product will be zero for orthogonal vectors. If your larger size is also changing the effective angle, this could account for the unexpected result.


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It's true that there are many inner products you can choose on $\mathbb{R}^3$. However, physics supplies the additional principle of rotational invariance: the result should not depend on our coordinate system. Now, any inner product of vectors $a$ and $b$ can be written as $$a \cdot b = a^T M b$$ for a matrix $M$. Rotational invariance tells us that $M$ ...


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The whole point in components is that when you add them, they must must give the original vector. The two components you've drawn don't. Their sum is not the original gravity vector. Remember that components are supposed to follow coordinate axes, so they are perpendicular to each other (in that way they take care of distinct directions so we can treat ...


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If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows : \begin{equation} \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...


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If we take your same two masses and write $m_1 = s \cdot m_2$, it's easy to understand what the scalar multiplication is doing: it's scaling. $m_1$ is $s$-times as big as $m_2$. In your language, multiplying $m_2$ by $s$ gives us a new mass which has the mass of $s$ [many] $m_2$'s. This may not seem to answer your question about work, because now the things ...



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