Tag Info

New answers tagged

0

There are a variety of ways to do this problem: a. Use the law of cosines to find the length of the third line. You will need to figure out what the interior angle between the two known lines is based on the angle each one makes with the ${0}^o$ direction. Then use the law of sines to find the angle. b. Find the $x$ components (parallel to ${0}^o$) and $y$ ...


-2

yes, find the point on the circle where the tangent pass trough (2,0,0), but be careful with the sign, because you will get two solutions that correspond to the two different directions of rotation (clockwise and counterclockwise). You must use the solution in the 4th quadrant that corresponds to the counterclockwise solution. the lower part of the ...


2

Suppose that the vector field $X$ is tangent to the hypersurface. Then $X\cdot v$ must vanish, since $v$ is a constant on the surface. But $X \cdot v$ is precisely what is meant by $dv(X)$. So the tangent space is precisely the null space of $dv$. Now since we have a metric, we can "raise the index", $$dv \mapsto g^\sharp dv = g^{\mu\nu} (dv)_\nu = V^\mu$$ ...


0

Curvature isn't really a problem -- find the normal vector at an arbitrary point, since then this is just a linear algebra problem. Lorentzian signature is also not a problem -- as long as we have an inner product, we're good. The procedure is analogous to the "3+1 split" of spacetime used, for instance, in the ADM formalism. Often we foliate spacetime ...


2

I don't think your confusion is related to curvature, and I think your question would persist even in the following flat space example. Consider a surface in Minkowski space $$\{(t,x,y,z): t,x,y,z, \in \mathbb{R}\},$$ with metric determined by $$\begin{bmatrix} -c^2 & 0 & 0 & 0 \\[0.3em] 0 & 1 & 0 & 0 \\[0.3em] 0 ...


0

If you have coordinates $(v,r,\theta,\phi)$ on $M$, and your hypersurfaces $\Sigma_v$ are characterized by $v=const$, then $$ \frac{\partial}{\partial r},\ \frac{\partial}{\partial \theta}\ \mathrm{and}\ \frac{\partial}{\partial \phi} $$ are coordinate basis fields on $\Sigma_v,$ and $\left.\partial/\partial v\right|_v$ is a vector field that pointwise ...


1

If you only had a finite number of rational numbers, you could add them a finite number of times, and multiply them a finite number of times and never worry about whether $\pi$ was a rational number. And in particular you wouldn't have to worry about whether there is a number that 3, 3.1, 3.14, ... etcetera converges to. You could do quantum mechanics the ...


0

A complete space means the idea of having one and only one representation for each vector $\mathbf{x}$ of space $\mathfrak{h}$, as $\mathbf{x} = \sum_i a_i \mathbf{e}_{i}$ ($i = 1, 2, ...$) with unique coefficients $a_i$ and orthonormal vectors $\mathbf{e}_i$ in $\mathfrak{h}$. In quantum mechanics, $\mathbf{e}_i$ will often be the eigenvectors of some ...


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 ...



Top 50 recent answers are included