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2

Short and a little inaccurate answer: vector is one-dimensional tensor, matrix is a two-dimensional tensor. More details now: Tensors are multidimensional arrays which have certain properties. Not every multidimensional array is a tensor, check this discussion for more details. There are two types of one-dimensional tensors: vectors and co-vectors. Both ...


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Your equation for the $\mathbf{v_{AB}}$, the velocity of $\mathcal{A}$ relative to $\mathcal{B}$ with the following formula is correct and general for all 'real' values of $v_{X_{i}}$ where $X=A, B$ and $i=x, y, z$. ...


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It is vector because it has a magnitude and a direction. Also typically the components of $\vec{\omega}$ are evaluated based on an inetial coordinate frame and thus only represent the motion of the body and not of the measuring frame. Rotational speed $\omega = \| \vec{\omega} \|$ Direction of rotation $\hat{z} = \frac{\vec{\omega}}{\omega}$ ...


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Yes, it is possible but it involves a lot more work. The problem is that the different components you want to separate are not similar enough for a simple subtraction to work. The frequencies are far from exactly the same, and even if they were approximately similar the relative phases would change quite fast. The phase information cannot therefore be used ...


1

A charged particle placed in a magnetic field experiences a force that causes it to deflect in the direction of force. Lorentz law is true. It will always be. The best way to apply the direction of the Lorentz force is by using Fleming's left (or right) hand rule. The right hand screw rule is helpful in analyzing the direction of curling of magnetic field ...


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Yes, the Lorentz force law holds, so whatever rule you're doing with your right hand must be wrong. All of these rules, in the end, come from the right hand cross product rule anyways. There are lots of things you can do with your right hand, though, so I wouldn't be surprised if one of them gave you the right direction.


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I'd say the premise is the other way around: We assume $\Delta A$ to be small, as we intend to make it infinitesimal later on, and so $\Delta \theta$ is also small. Also, when you look at $\Delta A\Delta \theta$, this is even smaller, or "quadratically small" so to say. When doing math with small deltas or infinitesimals, you usually only have to take the ...


4

It's because you have to replace the erroneous $\gamma$ by $\gamma^2$ (and similarly $m$ by $m^2$) in the inner product and because $$\gamma^2 m^2 c^2 - \gamma^2 m^2 v^2 = \gamma^2 m^2(c^2-v^2)=\dots$$ and $$\gamma^2 = \frac{1}{1-v^2/c^2} =\frac{c^2}{c^2-v^2} $$ and $c^2-v^2$ from the explicit factor cancels against the denominator of $\gamma^2$, while ...


1

What you have read is only valid in a vacuum. With air resistance the drag is a function of the total velocity, so in reality the deceleration on each axis also depends on the other.


2

Let the velocity at instant be v(vector) = (Vx) i + (Vy) j. i and j denote unit vectors, along x and y axis respectively. dv/dt = a(acceleration) = - gj. dv/dt = [(Vx(final) -Vx(initial))i + (Vy(final) - Vy(initial))j]/dt = - gj. By initial and final I mean Vx and Vy at time t and t + dt. As the resulatant is only along the y axis the X component must be 0 ...


2

Any volume integral of curl $\mathbf A$: $$ \int_V \nabla\times \mathbf A \,dV $$ can be calculated also as surface integral of $\mathbf A$: $$ \oint_\Sigma d\boldsymbol\Sigma\times\mathbf{A}; $$ here $\Sigma$ is boundary of the region and $d\boldsymbol \Sigma = \mathbf nd\Sigma$ is vector whose magnitude is that of area $d \Sigma$ has direction of ...


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The first green part is the Rodrigue's rotation formula. The second green part is a small angle approximation for $\delta \theta$.


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You can just use equation 3 as a definition. Yes, its dimension will differ from that of momentum, but this is not a big deal. That would not be a worse 4-vector than that of equation 1 - they differ just by a constant factor. And to use equation 2 you use a system of units where c=1, not 1 m/s.


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You have to think about what $\vec j,\mathrm{d}\vec l$ and $\mathrm{d}\vec s$ actually are: $\mathrm{d}\vec l$ points along the flow of the current. So does $\vec j$. So $\mathrm{d}\vec l$ and $\vec j$ are parallel, and indeed $$ (\vec j\cdot\mathrm{d}\vec s) \mathrm{d}\vec l = (\mathrm{d}\vec l\cdot\mathrm{d}\vec s)\vec j$$ holds in that case.


2

I use other symbols in order to prevent confusion in the following. Let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)\:$ as in above Figure. Then the volume charge density and the charge current density are expressed via Dirac $\:\delta$-function as follows \begin{align} \rho\left(\mathbf{x},t\right) & ...


2

As per, http://en.wikipedia.org/wiki/Four-velocity, we can define four-current density as: $J = \rho_0 U$, where $U$ is the four-velocity. Since it's a scalar times a four-vector, it's another four-vector. $$J = \gamma(v)(\rho_0 c,\rho_0 \vec{v})$$ $$J = (\gamma(v)\rho_0 c,\gamma(v)\rho_0 \vec{v})$$ Now it remains to show that this fits the definition you ...


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At every point the tangential direction is the unit vector of the velocity vector. If you have the velocity components $\boldsymbol{v} = (\dot{x}, \dot{y})$ at every instant, the you decompose this into a magnitude (speed $v$) and direction $\hat{\boldsymbol{e}}$ $$ \begin{align} v & = \sqrt{\dot{x}^2+\dot{y}^2} \\ \hat{\boldsymbol{e}} & = ...


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The easy way of doing this is to parametrice the trajectory. We have the cartesian definition, so let $ \textbf{r} : \mathbb{R} \rightarrow {\mathbb{R}}^{2} $ be: $$ \textbf{r}(t)=(t,f(t)) \quad\quad t\in (-\infty,\infty) $$ So we got the vector position as function of "time". Then, the velocity and acceleration vectors are defined by: $$ ...


1

The question is rather incomplete and confusing. By the way, it is used to consider surfaces as vectors when needed for computing surface integrals, like flux integrals, where the scalar product between a vector field $\vec A$ and a infinitesimal surface $\mathrm d\vec S$ is considered: $\vec A\cdot\mathrm d\vec S$. To this aim, the differential surface is ...


1

In special relativity there are two major assumptions: -the laws of physics are the same in all inertial frames -the speed of light that you observe is always the same, (thus independent of the relative motion between the light source and the observer). From this two assumptions follows the famous Lorentz transformations. In these Lorentz transformations ...


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The set of transformations that leaves the speed of light unchanged is the Lorentz group. Representation theory enables us to investigate the irreducible representations of the Lorentz group. The lowest-dimensional representations act on scalars four-vectors However, take note that usually we consider representations of the corresponding Lie algebra ...


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Rotation of a 3-vector We'll find an expression for the rotation of a vector $\mathbf{r}=(x_1,x_2,x_3)$ around an axis with unit vector $\mathbf{n}=(n_1,n_2,n_3)$ through an angle $\theta$, as shown in Figure . The vector $\mathbf{r}$ is analysed in two components \begin{equation} \mathbf{r}=\mathbf{r}_\|+\mathbf{r}_\bot \tag{01} \end{equation} ...


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How do we formally define vectors in physics? An excerpt from chapter one, page 12 of "Mathematics of Classical and Quantum Physics" Originally, we introduced a vector as an ordered triple of numbers. The rule for expressing the components of a vector in one coordinate system in terms of its components in another system tells us that if we ...


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A physical quantity is a vector if it transforms in the same way as a position vector when the coordinate system undergoes a transformation.


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Typically there are two kinds of transformation that do not change the outcome of situation. Think of a force vector $\vec{F}$ passing through a point $\vec{r}_A$. Any translation along the line of the force, in the direction $\vec{e} = \frac{\vec{F}}{\| \vec{F} \|}$ will not change the outcome. Any rotation about the line of the force would also not ...


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In general it changes although the reason is not exactly because its projections changes. For example. You start with a vector (let us say the electric field of a parallel plate capacitor) on the plane $xy$. Then you rotate the coordinate system by an angle. The components of the vector on the new coordinate system is changed. But the vector did not change ...


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Direction of a vector is determined by the components themselves. Now if the components are changed the direction gets changed by the above definition. All this is with respect to one reference frame.


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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. ...


1

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 : \begin{equation} \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 ...


1

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 ...


3

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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