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9

There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...


4

It's a vector. Instantaneous velocity $\vec v$ is defined as $$\vec v \equiv \lim_{\Delta t \rightarrow 0}\frac{\Delta \vec r}{\Delta t}.$$ In that equation, $\Delta \vec r$ is the displacement that occurs during time interval $\Delta t$. Putting on my math hat for physicists, the numerator is a vector, and the denominator is a scalar, so the resulting ...


3

It seems to me that it's not 'vectors' or 'vector algebra' which these students aren't grasping. Its the connection between a given 'physical phenomenon' and a corresponding 'mathematical representation'. I suspect this has something to do with conceptualising the physics rather than the mathematics. To put is simply: Physics $\neq$ Mathematics When ...


3

The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more. The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical ...


3

When one says that "kinetic energy is conserved in an elastic collision" that means that the total kinetic energy of the system of particles involved in the collision doesn't change. It does not mean that the kinetic energy of each particle is unchanged. For a two particle system, the kinetic energy of each will change, but the sum won't. Also, your ...


3

A general vector is written as $v = v^\mu \partial_\mu$. Its norm is defined as $$ g(v,v) = g_{\mu\nu}(\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(v^\mu\partial_\mu,v^\nu\partial_\nu) = g_{\mu\nu}v^\mu v^\nu (\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(\partial_\mu,\partial_\nu) = g_{\mu\nu}v^\mu v^\nu$$ where we have used linearity of the duals and ...


2

That's because you want $F_1 + \Delta F = F_2$ by additivity of vectors (for a more rigorous approach, see the formalization of affine spaces). Hence, $\Delta F = F_2 - F_1$ PS: I couldn't comment because of my low reputation, so I made an answer for so little


2

The notation $\nabla_1$ refers to the gradient with respect to the first coordinate $\mathbf{r}_1$. I think the most transparent way to do the derivation is to switch to the notation $\partial/\partial\mathbf{r}_1$, then expand the derivative using the multivariable chain rule, and then switch back to the nabla notation: $$\begin{align}\nabla_1 &\equiv ...


2

The planet will follow an elliptical path and as an ellipse is a 2D figure it can only be traced to a plane and from two points infinite planes can pass, so you need to decide first which plane you want which is the missing information, because of which you are not able to do so. If finally you form a plane $\vec r.\vec N=p$ then the direction of the ...


2

Restricting ourselves to just vector spaces without any extra structure, the theorem is true. One way to see this is to note that any member $f$ of the dual space is uniquely defined by the value it returns acting on the basis $\{\psi_n\}$, say $f(\psi_n) = z_n$ for complex numbers $z_n$. Then $V^*$ is isomorphic to $\mathbb{C}^\mathbb{N}$, the set of ...


2

I can't comment on the up-till-now, but here's something to try for the "now-and-henceforth". I suspect that many student difficulties have in the past been left unhelped by the fact that lecture courses were "linear" (no pun meant here): there was a set coursework and a set way of thinking about concepts that the lecturer or teacher chose that students ...


1

I have helped some school students through a course which was taught with Matter and Interactions which focussed on using physics principles to program computer simulations. Because of the programming, my students were comfortable with vectors from the first week. The computer made the vectors visual and as simple to manipulate as variables containing ...


1

I can't quite fathom the source of your confusion (I think it might have something to do with a focus on the notion of rotation here---angular momentum does not require rotational motion), so I'm having trouble writing a really clear response. For the moment I would rather offer a program for practicing the right skills rather than reinforcing the mistaken ...


1

The vertical component of the force doesnt do work, so a force with an obtuse angle can be considered to be oppossed to the diplacement. In any case, whenever work is negative, the system is losing energy.


1

You can calculate the work done by gravitational force as the product of its weight and y-displacement. If I have got your question right, the body is freely falling after the force tips it off the table. So the work done by your force will not be as you have written. It would've been correct if the force had been acting on the body throughout its ...


1

Careful between $\hat{r}$ and $\vec{r}$. The trajectory $\vec{r}(t)$ is a function that maps $t$ to a position vector $\vec{r}$. But the tangent to the trajectory is not the same as $\vec{r}$. Thus, $\vec{v}\cdot \vec{r}$ tells you nothing about the tangent vector. To be tangent to something means to be going in the same direction at exactly one point. The ...


1

Two issues: 1) the equation was derived assuming that the initial speeds were zero, so both masses started at rest. To get a more general expresion yuo need to integrate again (I'll check later if this is easily doable) 2) you can apply this equation in 3d, the equatiosn assumes that the masses start at rest and follow a staight path until they collide. ...


1

$F1 - F2$ is the same as turning $F2$ around 'head to tail' and you get $-\Delta F$ or a vector, which is the same magnitude, but the opposite direction to $\Delta F$. To get $\Delta F$ you need to reverse $F1$ as in the diagram I have modified from your diagram below


1

The answer is no, we cannot rule out forces not directed along the line joining particles with current theory. As you point out, magnetic monopoles are a counterexample, and, as CuriousOne points out, magnetic monopoles both fit into a classical framework, are consistent with the Standard Model and are actively, experimentally sought. Some historical ...


1

If $\theta$ is obtuse then $\cos\theta$ is negative and thus, $W=\vec{F} \cdot \vec{s}=Fs \cos\theta$ is negative.


1

The other answers are OK, but if I'm correct they are missing information. Firstly, to be completely thorough, a general approach to force questions is to split the forces into components as shown here. If you do that and add the vertical force components and the horizontal force components, you will get a net force. This net force is the direction of ...


1

Acceleration is simply a rate of change of velocity. So the magnitude tells you, how quickly velocity changes.



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