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15

You can use properties of the Levi-Civita tensor namely, $\epsilon_{kij}\epsilon_{kmn} = \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm}$ so that $\vec a \times (\vec b \times \vec c) = \hat e_i \epsilon_{ijk} a_j (\epsilon_{kmn} b_m c_n )$ $= \hat e_i ( \delta_{im}\delta_{jn} - \delta_{in}\delta_{jm} )a_j b_m c_n$ $= \hat e_i \delta_{im}\delta_{jn} a_j ... 15 Cross products can be understood from the perspective of Geometric Algebra, which defines the product of two vectors as the sum of a scalar and a 'bivector', a new sort of object that represents planes and areas in the same way that a vector represents lines and lengths. The scalar part is just the scalar product. Because the bivector part produces an object ... 14 In their modern form, vectors appeared late in the 19th century when Josiah Willard Gibbs and Oliver Heaviside (of the United States and Britain, respectively) independently developed vector analysis to express the new laws of electromagnetism discovered by the Scottish physicist James Clerk Maxwell. Quaternions were introduced by Hamilton in 1843. Important ... 13 There are several areas to which cross-product can be linked, including wedge products, axial vectors etc, but it is simple enough to be treated on its own. Below I will show that cross-product naturally arises if one seeks a bi-linear transformation of two 3d vectors, that gives rise to a third, perpendicular 3d vector. So let us define a quantity$\...

12

I always have trouble with this identity, so here's a fun way to derive it in three-dimensions. It may be argued that this method is a little convoluted, but I find it much easier to remember than the Levi-Civita contraction formula, and much less tedious than working out the components! Let's call the vector $\mathbf{a \times (b \times c) = d}$, and see ...

9

A cross product of two vectors actually is not a vector but a second rank antisymmetric tensor. In 3D this has 3 components so it is usually called an axial "vector". An axial vector is invariant under space inversion while a true vector changes sign. The antisymmetric tensor formed by two vectors is $$\begin{pmatrix} 0 & x_1 y_1 & x_1 z_1 \... 5 It's just a notation for vectors, like \vec{F} or \mathbf{F}. It's not that common, but I had at least one instructor use it for a class when I was in undergrad. If you look again carefully, you'll notice the work W is not underscored, because it's a scalar. 5 So you aren't doing the same thing as your teacher here. Notice how in what your teacher did they have \Vert d \mathbf r \Vert =ds, so they did not say that$$\frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {dr}{dt}\right)^2 $$this would not be true in say circular motion, where the position vector is changing but its magnitude is not. However, ... 4 You can easily see that for a stationnary circuit the curl of the electric field should be 0 (as you mentionned, this comes from the Maxwell-Faraday equation). However, this is not contradictory with the fact that \overrightarrow{E} seems to be circumferential. \overrightarrow{J} is indeed circumferential but \overrightarrow{E} is not. That's because ... 4 A more down-to-earth approach is to prove this identity in three-dimensional space by writing it down in terms of vector components: \vec{a} = (a_x, a_y, a_z), etc., and using the expression for the vector product$$ \vec{a}\times\vec{b} = \left| \begin{matrix} \hat{e}_x & \hat{e}_y & \hat{e}_z \\ a_x & a_y & a_z \\ b_x & b_y & ...

3

Reading your question I can think of two applications at the moment: When an electron moves through a magnetic field, the electron is deflected sideways. Empirically (through observations and measurements) it has been found that the direction of the deflection is always perpendicular to the surface spanned by the direction of the electron's movement and the ...

3

Yes, $\vec v=\frac{ds}{dt}\hat{t}$ So, $$d\vec{v}\cdot d\vec {v}=(d^2s\hat t +ds\,d\hat t)\cdot(d^2s\hat t+ds\,d\hat t)=(d^2s)^2+(ds)^2$$ But ${dv}^2=(d^2s)^2$

3

It is not a curl. This can be seen by expressing the curl in vector components. $$\nabla \times \mathbf M=\begin{pmatrix} \partial_yM_z-\partial_z M_y\\ \partial_zM_x-\partial_x M_z\\ \partial_xM_y-\partial_y M_x \end{pmatrix}$$ Here $\partial_x$ denotes the partial derivative with respect to $x$. The quantity $\partial_x\mathbf M$ is a vector just like $\... 3 Use the contraction identities of the Levi-Civita symbol. I strongly encourage you to prove these identities yourself as well, I think you will find it worth the effort in the long run. 3 In Figure-01 we have a solenoid of 1 turn. In case of uniform magnetic field$\mathbf{B}$parallel to the axis of the helical wire the flux through the helical surface is equal to the flux through its projection on the$xy-$plane that is the circle shown. This equality, which is exact and not approximate, is proved in the Figure-03 below. In Figure-02 we ... 2 First lets be on the same page regarding what is angular velocity ? Angular velocity often, denoted as$\omega$is the rate of angular displacement, denoted as$\thetawith respect to time i.e. you might have seen this equation a lot $$\displaystyle{\vec{\omega} = \frac{\vec{\theta}}{t}}$$ and if we are talking about instantaneous angular velocity then : \... 2 There is in fact a way of expressing angular velocity in such a way that there is no ambiguity of what-part-of-this-is-convention. Angular velocity occurs in a plane, and it has direction and magnitude. To specify a plane two vectors that lie in that plane are specified, with the order of the two vectors giving the direction of the angular velocity. The ... 2 Assuming constant electric field we've \begin{align} \vec{T} &=\int \vec{r} \times(\rho(\vec{r}) \vec{E} d \tau) \\ &=\int(\vec{r} \rho(\vec{r}) d \tau \times E) \\ &=\left(\int \vec{r} \rho(\vec{r}) d \tau\right) \times \vec{E} \\ &=\vec{p} \times \vec{E} \end{align} 2 Why is the component of contact force is not like F_{contact} \cos \theta and F_{contact} \sin \theta but here they have no trigonometric values like we normally make components for a force. Like F \cos \theta and F \sin \theta. They are still the components of the contact force and that's why you get a relation between the normal force and the ... 2 I assume that you'd have no difficulty working out the flux linked with a closed circular loop if the flux density is uniform and normal to the plane containing the loop. With a long solenoid (a helix of small pitch) of n turns we usually treat each turn as a circular loop having, in the case you present, a flux \Phi=\vec B. \vec A =BA through it. The ... 2 If you are dealing with a composite system as it seems, you don’t need to change the order of \psi and \phi from (1) to (2), since the first (second) ket/bra always refers to the first (second) subsystem of your larger system. Along the same reasoning, you don’t need to change the order if the two kets/bras refer to two different degrees of freedom of the ... 1 How is that possible? It's only possible because we consider 3 dimensions. It wouldn't be possible on a 2D-plane for example. If you have 2 non-collinear vectors in 2 dimensions: you couldn't find a third vector perpendicular to both of them. You're stuck inside the plane defined by those 2 vectors. But if you consider 3 dimensions, it's perfectly possible ... 1 You can only say that PxAC=QxBC if you are told that there is no net torque (no rotation) of the rigid body about the point C. Of that is the case then the resultant of the two forces P and Q at at point C is simply a vertical force of R=P+Q with no rotation about the point C. Let us say that there is no torque. How do i prove then P . AC = Q . BC ? If ... 1 The definition for any steady current distribution is {\bf m}= \frac 12 \int {\bf r}\times {\bf J}\, d^3x. $$This quantity is independent of the origin chosen for {\bf r} because$$ \int {\bf J} \, d^3x={\bf 0} $$for a steady current that obeys \nabla \cdot {\bf J}=0. I leave it up to you to specialize this general case to one of a current ... 1 The given solution integrates forces in the direction of the chain such that constraint forces from the frictionless tube can be ignored. Your solution does give the force due to gravity, but lacks the normal force from the frictionless tube, so it gives the incorrect net force. 1 Use the individual contributions of each force, to sum up the combined moment.$$ \begin{aligned} \vec{F}_{R} & = \sum_i \vec{F}_i \\ \vec{M}_{R_O} & = \sum_i \vec{r}_i \times \vec{F}_i \end{aligned} \tag{1}$$Now you can where the resultant force's line of action is in 3D.$$ \vec{r}_{R_O} = \frac{ \vec{F}_{R} \times \vec{M}_{R_O} }{ \| \vec{F}_{R}... 1 Each force\vec {F_i}$exerts a moment$\vec {M_i}$about$O$. You can add the moments of the individual forces just as if they were vectors (technically they are pseudovectors but they can still be added like vectors). So the total moment$\vec{M_{R_0}}$is just the vector such of the individual moments i.e.$\displaystyle \vec{M_{R_0}} = \sum_i \vec {M_i}$... 1 Since the question mentioned higher-dimensional space, I wanted to give an answer that works in any-dimensional space, not just 3. I'll start with formal, mathematical definitions and then connect them to physical intuition. Rotations in$n\$-dimensional space form a group. Specifically, they form a group called the special orthogonal group, which is denoted ...

1

I am learning about Angular velocity. And I am confused that is the direction of angular velocity just a definition or has a physical significance. You are going to get confusing answers, because your question as stated doesn't mean much. But it means something.... There are things in mathematical notation that are basicly arbitrary. Somebody chose to write ...

1

The vectors representing rotation are chosen along the axis if rotation because that is the only direction in the system which is usually not continuously changing direction. That said, such vectors can accurately represent the direction and magnitude of rotational quantities.

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