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17

You are correct. Position is a vector when you are working in a vector space, since, well, it is a vector space. Even then, if you use a nonlinear coordinate system, the coordinates of a point expressed in that coordinate system will not behave as a vector, since a nonlinear coordinate system is basically a nonlinear map from the vector space to ...


4

Great reasoning: as in Uldreth's fantastic answer but I would add one more thing that may help cement your good understanding in place. Co-ordinates are absolutely not vectors, they are labels on charts and are no more vectors than your street address is a vector. Almost certainly the reason people make the implication that you have correctly identified as ...


4

To my mind, Newton's equation makes the most sense as an equation of vector fields. Let $(M,g)$ be a (Pseudo-)Riemannian manifold with Riemannian connection $\nabla$. Then the equations of motion for a position-dependent conservative force $F$ are given by $$m {}^{\gamma}\nabla_{\frac{d}{dt}} \frac{d\gamma}{dt}=F\circ \gamma=-(\nabla V) \circ \gamma,$$ where ...


4

It does give "Coulomb's law" with $\frac{1}{r^3}$, it gives it in its proper vectorial form $$ \vec E \propto \frac{\vec r}{r^3}$$ which, when taking the absolute values, yields the form you are probably more familiar with $$ E \propto \frac{1}{r^2}$$ since $\lvert \vec r \rvert = r$.


3

Notation: I will write a Poincaré transformation as ${x'}^\mu = {\Lambda^\mu}_\nu x^\nu + a^\mu$, the operator representing this transformation on the Hilbert space is $U(\Lambda, a)$. An infinitesimal transformation with ${\Lambda^\mu}_\nu = \delta^\mu_\nu + {\omega^\mu}_\nu$ and $a^\mu = \epsilon^\mu$ can be expanded as $$ U(\delta + \omega, \epsilon) = 1 ...


2

Tension is an internal force in a body, such as a rope, that resists any attempt to pull the rope apart. Simply, tension arises due to intermolecular interactions, and if it did not exist, ropes would fall apart the moment you pull on them. Now, it is necessary to distinguish between internal and external forces for a body. External forces are forces that ...


2

Here is a bare bones easy way to see that coordinate tuples are not 4-vectors. Start in an inertial coordinate system in flat spacetime. Change the coordinate system with a constant translation: $x' = x + A $ $y' = y$ $z' = z$ $t' = t$ Even in this idealistic case, 4-vectors and coordinate tuples transform differently. The components of the 4-vectors ...


1

You know from the question that the forces are in the same direction as the sides of the equilateral triangle, so they are at 60 degrees to each other. They all have different magnitudes so cannot possibly form an equilateral triangle if placed "end-to-end". To work out the answer, you can calculate the horizontal and vertical components of each of the ...


1

The equilateral triangle only tells you the directions of the vectors. The vectors must be ``floated'' so that they all act on the point. Each vector has its own length. Here's the diagram:


1

In ordinary vector spaces, the dot product $\cdot$ is a binary operator which takes a pair of vectors $(A,B)$ in the space to the field over which the space is defined. Formally, for a vector space $V$ over a field $K$, the dot product $(\ \ , \ )$ is a bilinear map $$(\ \ , \ ): V \times V \to K.$$ The inner product only has assumes the standard meaning ...


1

The introductory quote defines $\tau$ and $\nu$ to be unit vectors. That means that their magnitude is 1. $\mathbb{i}$ and $\mathbb{j}$ are unit vectors along $x$ and $y$ axes. That's a standard notation. Some of the comments are stressing the fact that the basic understanding of vectors is missing. I have a feeling that you've jumped into a 'complicated' ...


1

$v$ isn't referring to either $v_1$ or $v_2$, necessarily; $v_1$ is representing the vector before it moves, and $v_2$ is the vector after this movement. If we are working in polar coordinates (the reason he is using $v_\perp$ and $v_\parallel$), then let's suppose this small movement isn't changing the magnitude of the vector, it is just changing the ...


1

You can write $$\text dV = - m v \text d v$$ where $v$ is the velocity.


1

The tension on a string between two objects (note: your top diagram shows t̲h̲r̲e̲e̲ objects) is analogous to the force between two objects elastically colliding. The force exerted by the one end of the string is opposite and equal to the force exerted by the other end of the string; both forces must be parallel to the string and pointed towards its center. ...


1

We treat the string/rope like another object. This object exerts forces on other objects such as the hanging mass (in your picture). However, a string, by its very nature, can never "push" another object, it can only "pull" another object. That "pull" is a force which we give the name tension. Thus, tension will point away from the mass in the direction of ...


1

A representation of the Lorentz group implies a set of matrices $(S_{\mu\nu})_a{}^b$ (one for each $\mu,\nu$) that satisfy the Lorentz algebra, i.e. satisfies a relation of the form (I am not keeping track of signs) $$ [S_{\mu\nu} , S_{\rho\sigma} ] = i ( \eta_{\mu\rho} S_{\nu\sigma} + \cdots ) $$ What this means is that there is a vector space, with vectors ...


1

Your strategy is fine (though I haven't carefully checked if your rotations are correct). Since rotations don't commute, you need to be careful to apply them in the correct order. You say that your rotation about the x-axis, let's call it $R$, followed by your rotation around the z-axis (be careful whether it's around the old or the new z-axis!), let's call ...



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