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You can also see it in the context of group representation theory. $O(n)$ is generated by the same infinitesimal generators as $SO(n)$ plus the discrete generator $\mathbf P$ which is parity. The parity operator generates a discrete subgroup $\mathbb Z/2\mathbb Z$ in $O(n)$). So if we choose a representation $\rho_{SO}$ of $SO(n)$ and a representation ...


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First of all, as this is a homework question, I can't tell you the complete solution. Choose P as the origin of coordinate system and resolve the forces into x and y component. And as the body is in equilibrium, the net force is zero. So you get these two relations (when the net force on x and y component equated to zero.) $$G\cos\theta=H\cos\phi\\ ...


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A vector is a basic mathematical construct. There can be many types of vectors (velocity, position, force, etc...). Your first chapter is defining a vector in general. The second is introducing a vector that describes position. If you take the difference of two position vectors, it is a displacement vector. Thus. Through algebra the difference of a ...


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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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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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Since the wavelength is the distance between the two consecutive crests or troughs it isn't a vector quantity. It is scalar. Simillarly, Frequency is a scalar quantity, since it just number of crest and trough per second. How can such number have direction.


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Vectors and vector notation can tend to simplify equations and algebra. They might feel like they just complicate things at first, but once you get used to them they can provide an intuitive way to simplify calculations and concepts. Consider for example Newton's second law. While without vectors we'd have to write $$F_x = m \times a_x$$ $$F_y = m \times ...


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What would your "single variable" for direction look like? You'd need two angles, or something equivalent, to describe it, and two angles and a magnitude is exactly a vector in spherical coordinates.


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Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are.


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By convention, North and South rank higher in the English language. Thus we have Northeast and not Eastnorth. You have presumed a flat earth. If you start 1.5 km from the North Pole, then after skiing 3 km east and 1.5 km north, you will be at the North Pole, 1.5 km North from your starting point.


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You asked "how do I sum more than two vectors". In general, you can write your vectors as $(x_i, y_i)$ where $i$ counts from 1 to 4 in your case. Then the sum vector is $$x_{sum} = \sum x_i\\ y_{sum} = \sum y_i$$ The forces you have are all pointing towards the center, which means that for each of the forces, you will have either $x_i=y_i$ or $x_i = - ...


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It's hard to picture the things you want because you're working in an infinite-dimensional space, and I don't know anybody who can intuitively 'see' in that kind of space. Instead, let's consider a finite-dimensional space: the polarization of a massless spin 1 particle traveling along the $z$ axis. The polarization state is simply described by a unit ...


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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 ...


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$\newcommand{\real}{\mathbb R}\newcommand{\field}{\mathbb F}\newcommand{\cx}{\mathbb C}\newcommand{\ip}[2]{\left< #1,#2\right>}$We need to dive into mathematics of vector spaces and inner products in order to understand what a vector means and what is it mean to take a scalar product of two vectors. There is a long post ahead so bear with me even ...


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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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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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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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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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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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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 ...


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It all depends on the direction you assign while using $dx$. It is generally a vector, so the sign will automatically be adjusted according the operation used and its direction relative to other operand vectors. In your case, you are calculating the work done by the spring. The force due to a spring is always directed to its mean position. With this in ...


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basic concept is to set up a coordinate system , Cartesian would be good here Here, I'm calculating work done BY the spring origin is the point where spring is in natural position (for us to use F = -Kx , otherwise, it would also contain some constant too, ex- if our spring's natural length is at (2,5) and spring only moves in along x axis direction, ...


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You might get a wrong sign due to $cos(0)= 1$ and $cos(\pi)= -1$. You either take $x$ as a cartesian coordinate and integrate over it or you use polar coordinates with some maximum elongation $x$ which is constant. In your question you are mixing up both. The conventional sign here is $-$, however it depends on your way of counting the work, ie work done to ...


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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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friction always tries to oppose the relative motion ... or you can also choose any arbitrary direction if your answer shows negative sign than assumed direction is the opposite direction of real friction force acting ...


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It is pretty much simply a short way to notate both vector field operations by looking at $\nabla$ as a vector operator by writing \begin{equation} \nabla=\left(\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right) \end{equation} in $\mathbb{R}^3$, or equivalently \begin{equation} \nabla=\frac{\partial}{\partial ...


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One way to see that $\vec{A}(t)$ rotates around $\vec{\alpha}$ is that its tangent (or derivative) is always perpendicular to $\vec{\alpha}$ and $\vec{A}$. In order to give more definitive proof it might be easier to write the differential equation is matrix form $$ \begin{bmatrix} \dot{A}_x \\ \dot{A}_y \\ \dot{A}_z \end{bmatrix} = \begin{bmatrix} 0 & ...


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The equation is quite relevant in physics. Think of the precession of a magnetic dipole in a magnetic field, NMR etc. \begin{eqnarray*} \frac{d\mathbf{A}}{dt} &=&\mathbf{\alpha }\times \mathbf{A} \\ \frac{d\mathbf{\alpha \cdot A}}{dt} &=&\mathbf{\alpha \cdot \alpha }\times \mathbf{A}=0\Rightarrow \mathbf{\alpha \cdot A}\;\mathrm{const} \\ ...


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You start off by drawing a free body diagram, showing the x and y components of the force. Then you write a force balance for the x direction and the y direction to get the accelerations in these directions. Then you can determine the kinematics of the motion, once you know the acceleration components.


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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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Apply: \begin{equation} F = ma \end{equation} to your particle to get a pair of acceleration vectors. You can then use the acceleration to work out the particle position at a given time. \begin{equation} x = x1 + at^2/2 \end{equation} If you meant impulse instead of force then the process is similar but you equate the impulse to the total change in momentum ...


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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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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 ...



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