# Tag Info

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The state is a vector in the Hilbert space of the Hamiltonian, which gives it a natural basis in terms of the eigenvectors; distinct eigenvalues then exist in distinct (orthogonal) subspaces - for degenerate values the subspaces are larger, but they are still distinct from all others. Clearly this situation gives many advantages in analysis. However, this ...

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This is an interesting question and the answers which have been given show that the $v$ in your equation should be called the magnitude of the velocity or just the speed of the wave. The mixing of the terms speed and velocity happens all the time. Now there is an equation for wave velocity but in comes about in a somewhat convoluted way. Suppose that you ...

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This is one of those things that (intentionally) gets conflated, though it may be better if we were more consistent about keeping them separate. So, points don't form a vector space. It makes no sense to ask "what's the location of New York plus the location of DC". However, given two points we can subtract them and get a displacement, and we can add that ...

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Suppose $\left\{\left|e_i\right\rangle|i\in I\right\}$ is an orthonormal basis of a Hilbert space $\mathcal{H}$, viz. $\left\langle e_i |e_j\right\rangle =\delta_{ij}$. Then the identity operator from $\mathcal{H}$ to $\mathcal{H}$ can be written as an outer product $$\mathbb{I}=\Sigma_{i\in I}\left|e_i\right\rangle\left\langle ... 2 Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are. 2 If we differentiate the definition of radians, l = R\theta, with respect to time we get v = \dot R\theta + R \dot\theta. As R is constant, \dot R = 0, so we have v = R \dot\theta. However v, R and \dot\theta are scalars, whereas you want the vector. So multiply by the tangent unit vector, \hat{T}, and we arrive at \vec{v} = R \dot\theta ... 2 It is pretty much simply a short way to notate both vector field operations by looking at \nabla as a vector operator by writing $$\nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right)$$ in \mathbb{R}^3, or equivalently \nabla=\frac{\partial}{\partial ... 2 The dx is actually a vector d\vec{x} (because \vec{x} is actually a vector). A vector's sign depends on your coordinate axes, e.g. if you pick right to be positive then left pointing vectors are negative. In general, the work done W is given by$$ W = \int \vec{F} \cdot d\vec{x} = \int Fdx \cos(\theta) $$where \theta is the angle between ... 1 The concept you're after is the dot product between 2 vectors (your displacements). More specifically you want to use$$ \vec{a} \cdot \vec{b} = |a| |b| \cos(\theta)  to find $\theta$.

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Describing a rotation as a vector, with the direction of the vector along the axis of rotation, and the magnitude of the vector as the angle, is known as the axis–angle representation.

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Can an angle be defined as a vector? It depends on what you mean by "vector". If by "vector" you only mean something that has a magnitude and a direction, then yes, the axis-angle representation qualifies as a "vector". To a mathematician, a vector is something that is a member of a vector space. In this context, the axis-angle representation fails to ...

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Note that pendulum motion is only sinusoidal for small angular displacements; as you increase the amount of swing the harmonic approximation fails. Lagrangian mechanics gives you a handle on all of the cases.

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This is a math problem and is probably better suited for Mathematics SE. However, I will spare some tips on solving this problem: Parameterize the curve as l = (x(t),y(t)) and solve for dl. (hint: let one variable equal t and solve for the coordinate. You need x and y as functions of t, hence parameterization.) You need to input the parameterization into ...

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No - in the video, the instructor calculates $A$ and $B$ which are the total tension in the cables. Taking a screen shot at 6:16 shows this: The horizontal components are $A \cos 60$ and $B\cos 40$ respectively - and these cancel exactly. $A$ and $B$ are indeed the total tension. You could get that by summing the $x$ and $y$ components according to ...

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First of all you have to consider that applying a force to a particle can make it move, it depends on the boundary conditions like the presence of friction. According to the data you have provided you are in the very simple case of just the particle and the force acting on it so assuming it has mass $m$ (this data is needed) hence the particle will move ...

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