# Tag Info

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Products are omnipresent in physics – and even in less quantitative sciences – from Day One. The product $A=XY$ of two numbers may be visualized as the area $A$ of the rectangle whose sides are the two factors $X$ and $Y$. When embedded in physics, this simple formula $A=XY$ already includes units: the sides are in meters and the area is in squared meters. ...

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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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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 : \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...

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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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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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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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A dot product is defined as the product of two vector amplitudes times the cosine of the angle between them. In one dimension, that 'cosine' term is either $1$ or $-1$. The RMS average is $1$. In two dimensions, it has an RMS average value of $1/\sqrt{2}$. This generalizes, and between two $N$-dimensional vectors, the RMS average value of the cosine is ...

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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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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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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 ... 2 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 ... 1 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$M$... 3 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, ... 0 Once you have the metric in u,v coordinates, you notice that$g_{uv}=0$and this shows that the basis vectors$e_u$and$e_v$, respectively tangent to the v=const and u=const lines (no typo, you see why?) are orthogonal. 0 The cross product of two vectors is the area of a parallelogram defined by those vectors. If the angle is 90 degrees then the cross product is simply the magnitude of vector 1 multiplied by the magnitude of vector 2 (i.e are of a rectangle). another way of writing the cross product is |v||u|sin(theta). If theta is 0 (both pointing in same direction) or 180 ... 0 A vector may be thought of as a directed line segment so that it has a magnitude (the size of the vector) and a direction. It is conventional (and almost common sense) to take the size of the vector to be a positive quantity. When you talk about the components you are using of the of vector you are asking for the size of that vector in some direction. For ... 1 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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A particle following a prescribed path has its velocity vector parameterized as $$\vec{v} = \vec{e} \,v$$ where $\vec{e}$ is the tangent vector and $v$ is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector $\vec{r}$. Use $\vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} ... 2 These two formulae must give the same result. I think there are some mistakes in your calculation. The result a) is correct. a)$(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 $$... 1 To add to Dilithium Matrix's answer, to simplify, you can look at v_1 and v_2 as any vectors, not necessarily velocity vectors. We know that if a = b and b = c, then a = c. Hence, to say that \theta = \beta is to say that \alpha = \beta. You can then use geometry to show that there is only one other case where, for O' \neq O, \alpha will ... 2 No. Trivial counterexample would be if the origin, o, is just below the tangent point for v1, directly below it. In that case, beta clearly can be 90 degrees or more, even if v1 and v2 are actually in the same direction! 0 This diagram shows part of a surface cut up into small elements (shaded pink) of area dA. The normal to the area is the vector \vec {dA} and is of magnitude dA. What is required is the projection of the magnetic field \vec B onto the vector \vec{dA} which is B_\bot = B \cos \phi in the diagram. To do this the dot product is taken \vec B ... 0$$ \int \mathbf{B} \cdot \,d\mathbf{A} $$just means$$ \int B\,dA\,\cos\theta. $$d\mathbf{A} points in the direction perpendicular to the surface, so if this is in the xy plane, it'll be in \mathbf{\hat{z}}. I.e. your usual integration over the area ("integrand") weighted by a factor of \cos\theta. \theta is the angle with the z axis. Keeping ... 0 The integration variable is the surface vector: flux is an oriented thing. It is the "normal" integration of the scalar product of the field and the surface vector:$$\Phi=\iint_S \vec{B}\cdot\vec{\mathrm dS}$$Or, if \vec{n} is the normal to the surface at the point of integration:$$\Phi=\iint_S \vec{B}\cdot\vec{n}\ \mathrm dS$$0 Physically, a tensor is just a collection of vectors that physicists study as a single object because the vectors are related to one another in a physically "interesting" way. Tensors don't possess additional properties to the ideas of magnitude and orientation of a vector; all that can be done is to assign a vector to certain arrangements of vectors of ... 0 On dot product you get magnitude, in units of product of operands. On cross product you get vectors with direction , in units of product of operands. 0 The initial answer is exactly right, and your response shows you understand that the dot product of the force and velocity will give you a scaler value that leads to the right answer. However when I read the question, I was impressed at how hard it is to visualize what is really going on, and the incurable teacher in me thought you might be interested in ... 0 Joules per seconds is Power. Power is work done per unit time. (J/s) Work is Force x Distance x Cos(angle) Dividing by time to get power yields Force x Distance / Time x Cos(angle) Velocity is Distance / Time and therefore Power = Force x Velocity x Cos(angle) Since the direction of the application of the force doesn't matter in this problem you can ... 0 It's just asking to find a vector such that$\vec{v} + \vec{v} = \vec{v}$. We get the equation$2\vec{v} = \vec{v}$which is only satisfied for, as you said,$\vec{v} = 0$, the null vector. Edit: Well more accurately, as it's talking about magnitudes,$|2\vec{v}| = |\vec{v}|$, but the point still stands. 0 The object is not in equilibrium. However, you can still calculate$F\$. You know that the string isn't lengthening or shortening, so the net force in the direction of the string is zero. So you can decompose the three forces in the problem into components along the string and components perpendicular to the string, and set the net force along the string to ...

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