Tag Info

New answers tagged

1

Vectors can be defined in multiple different ways. Here I will show the four ways and explain how they are equivalent$^1$. An equivalence class of curves. Given a curve parameter $t$, curves are considered equivalent if they have equal zeroth- and first-order derivatives at $t=0$. In other words, a vector at a point $p$ is an equivalence class $[\gamma]$ ...


1

When physicists say that a vector is an n-tuple that transforms according to... they expect you to guess a lot that isn't said. What they mean is that for each basis you are given an n-tuple of numbers. And when you take the matrix that gives you the change of bases for any two bases and you apply the formula in their "definition" the first n-tuple goes to ...


0

From a strictly mathematical point of view, the reason why the notion of: $$\text{“$n$ globally defined scalar fields $f_i\colon M \to \mathbb R $”}$$ is different from the one of: $$\text{“a vector field $X\colon M \to TM$”}$$ is indicated in Qmechanic's answer. The notion of vector field is intrinsic, that is, it doesn't depend from the choice of a ...


2

OP wrote (v3): Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go. It is (implicitly) implied by OP's notation that the component functions ...


0

There's been some good discussions linked and 0celo7 wrote the answer with great brevity, but I'll take a crack at adding some exposition. Let's consider situations of rotation: We can consider clockwise rotation (for our cartesian basis $\{e_i\}$) for some vector $V$, $ \hat x' = \hat x\cos(\theta)+\hat y\sin(\theta) $ $ \hat y' = -\hat x\sin(\theta)+\hat ...


0

I think that the physicists and mathematicians in this situation take two different ways to obtain the same thing. As Borges would have said, everyone is either platonist or aristotelian; in this case (and probably always) the mathematicians being the former and the physicists the latter. From the mathematical point of view, you the general structure ...


0

Take Cartesian coordinates for the real plane and transform them into polar. Does the set of coordinates $(x,y)$ transform as a vector? If you work out this example you'll see that this transformation, unlike linear ones, doesn't involve the Jacobian of the transformation. In the linear case this is just an accident.


0

Dynamic pressure means orderly motion. Magnitude and direction are involved. With more than one variable it becomes a vector.


1

Yes , the normal to the surface is the direction of reaction force. And the direction doesnt depend on the material of the object . But note that if friction is considered , direction of net reaction force changes


0

The same vector x⃗ can be written in terms of the basis vectors ei of the dual space V* as: xiei, why is this true? It's not true. The elements of the dual space are not vectors as we ordinarily conceive of them geometrically, e.g., directed line segments. Rather, they are (geometrically) a set of oriented surfaces with a density proportional to ...


2

If $V$ is a (finite dimensional) vector space with no additional structure, then $V^*$ is a different vector space of the same dimension, hence isomorphic to $V$ --- but it would be a great mistake to think of $V$ and $V^*$ as "the same'' or even ``almost the same'', because there is no preferred isomorphism between them. In other words, you can pick a ...


0

You can resolve the E vectors coming out of the yellow surface into a component parallel to the surface and a component perpendicular to the surface: $E = E_{parallel} + E_{perpendicular}$. As I say to my students a lot of the time, draw the triangle! The triangle here has a hypotenuse of one of the E vectors (in red), one leg of the triangle is the ...


0

There are simple examples, and quite complex examples. You could plot stars, with a vector in the direction you see the starlight from, and a length telling you the brightness. You are simply using a 3d vector to represent 3 bits of information (two for the direction, one for the brightness), and I don't know any way to add them sensibly. Another example ...


0

1.So you have a position vector given by $$\vec{r}=\left(8t-4.5t^3\right)\hat{i}+\left(10-2t^2\right)\hat{j}$$ If you use the definition of velocity as $$\vec{v}=\frac{d\vec{r}}{dt}$$ then we differentiate each component of the vector to get the velocity vector. $$\vec{v}=\left(8-13.5t^2\right)\hat{i}+\left(-4t^2\right)\hat{j}$$ Acceleration $\vec{a}$ is ...


3

This is a common misconception. When you apply a force upward on the object, the "reaction" force in Newton's 3rd Law is NOT the force of gravity down on the object; they do not have to be equal, and as you said, cannot be equal if you are to accelerate the object upwards. It is just a confusing coincidence that the force of gravity kind of looks like a ...


1

First, angular momentum isn't measured about an axis. It's measured about a point. Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in. Now, about ...


6

Comments to the question (v3): I) The notions of vectors, tensors, scalars, etc, depend on contexts in physics, cf. e.g. this and this Phys.SE posts and links therein. II) In OP's context, these notions refer to representations $\rho$ of the Lie group $SO(3)$ [and the corresponding Lie algebra $so(3)$] of 3D rotations, cf. e.g. Ref. 1. Let $\mathrm{i}L_k$, ...


3

The Poisson bracket you wrote only works for position (which is not a vector in general, as coordinates do not transform as vectors or any other tensor quantity). For general vectors the correct Poisson bracket is $$\{L_i,A_m\} = - (\mathbf p\times\nabla_{\mathbf p} + \mathbf r\times\nabla)_i A_m,$$ which reduces to the relation you wrote if you take $A_m = ...


1

One way to prove it would be simply just do it the hard-way. Just find out the resolving vector. Lets assume there is a vector $\vec{C}$ which need to be resolved into vectors at direction $\alpha$ clockwise and $\beta$ counterclockwise to the Original vectors. Lets identify these directions by $\hat{A}_\alpha$ and $\hat{B}_\beta$ respectively. $$ i.e.\ \ ...


1

Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, ...



Top 50 recent answers are included